### Create a StudySoup account

#### Be part of our community, it's free to join!

Already have a StudySoup account? Login here

# ComplexVariables MATH322

Drexel

GPA 3.63

### View Full Document

## 70

## 0

## Popular in Course

## Popular in Mathematics (M)

This 177 page Class Notes was uploaded by Moshe Swift III on Wednesday September 23, 2015. The Class Notes belongs to MATH322 at Drexel University taught by HugoWoerdeman in Fall. Since its upload, it has received 70 views. For similar materials see /class/212306/math322-drexel-university in Mathematics (M) at Drexel University.

## Reviews for ComplexVariables

### What is Karma?

#### Karma is the currency of StudySoup.

#### You can buy or earn more Karma at anytime and redeem it for class notes, study guides, flashcards, and more!

Date Created: 09/23/15

Student Solutions Manual for use with CompleX WVariables and Applications Seventh Edition Selected Solutions to Exercises in Chapters 17 by James Ward Brown Professor of Mathematics The University of Michigan Dearborn Ruel V Churchill Late Professor of Mathematics The University of Michigan Higher Education Boston Burr Ridge IL DubuquelA MadisonWI New York San Francisco St Louis Bangkok BogOta Caracas Kuala Lumpur Lisbon London Madrid Mexico City Milan Montreal New Delhi Santiago Seoul Singapore Sydney Taipei Toronto Table of Contents Chapter 1 1 Chapter 2 22 Chapter 3 35 Chapter 4 53 Chapter 5 75 Chapter 6 94 Chapter 7 118 12 quotCOMPLEX VARIABLES AND APPLICATIONSquot 7e by Brown and Churchill Chapter 1 SECTION 2 1 a fi i i1 i sEi i w5 2i b 232 1 4 36 2 1 8 1 1 l i c 313 1 E 100510 21 2 a Reiz Reix iy Re y ix y 1m z b Imiz Imix iy Imy ix x Rez 3 1z21zlz1z11zzl1zzlz 1zzz2 12222 4 If 2411 then 22222lii2 21li2t2i 2F2i20 5 To prove that multiplication is commutative write 2132 x1 y1x2y2x1x2 ylyz Yixz xly2 xle quot 3 23 1r 3 27 1 352371 xz hxxvh 3251 6 a To verify the associative law for addition write 21zzza 11 1xz Y2x3 3 x1 x2 1 2x3 3 x1 x2x39 Y1 2 Y3x1x2 353 Y1 2 Y3D x1 Y1x2 x3 3 2 135 xivY1x2 2x3 Y3I Z1Zz 23 10 b To verify the distributive law write ZZl 22 xYx1 J 1x2 2 x x1 x2 13 2 9051 132 yyl VY2 yxl yx2 75 992 xxl quotYYI xx2 39D y yxl 4393 yi yx2 302 xx1 yyp yx11xy1xx2 yy2 W H xyx1y1xyxpy2 zzl 222 The problem here is to solve the equation z2 2 1 0 for z x y by writing xJXxJ LY 10 00 Since x2 y2 x12xyy0o itfollowsthat xz y2xl0 and 2xyy0 By writing the second of these equations as 2x 1y 0 we see that either 2x1 O or y0 If y0 the first equation becomes x2x10 which has no real roots according to the quadratic formula Hence 2x 1 0 or x 12 In that case the rst equation reveals that y2 34 or y i Z Thus 1 V5 zxy i7 SECTION 3 1 a 12i 2 i l2i34i 2i 5i 510i 5 10ig 39 34139 5139 3 4i34i Six Si 25 25 0 Si Si 5139 1 l i2 i3 i 1 3064 TE 6 1D4 1i1iz 41392 4 1z z since z lz zl l 20 0 1 L z 20 E z lz 3 2122 zsz4 21 22z3z4 zlzzz3z4 Z1zgz2z4 21 23 2224 2113x2224 6 2122 2611 21i22l 23 0 2 0 Z324 2324 Z3 Z4 23 Z4 23 24 7 H5izlilltu gtW5 zzz 22 z 22 z 22 Z 22 SECTION 4 10 2 a z12i z2 l 21 22 51 a 22 c z31 Zz14 n9 y 72 Zn 0 x zi zz d ZixiiYi9 z2x1i 1 y 11 5 2 O Zlz2 x 2 Inequalities 3 Sec 4 are Rez S lRezl S Izl and Imz S lIsz S lzl These are obvious if we write them as xSlxISxJci wy2 and ySlyISylx2y2 3 In order to verify the inequality wElzl 2 IRezI lIsz we rewrite it in the following ways JEWZ xlyl 2x2 y2 2x2 2xly ylz lxlz 2xy IyIZZ 0 ext Imz 20 This last form of the inequality to be veri ed is obviously true since the lefthand side is a perfect square 4 a Rewrite lz 1 i 1 as Iz l i 1 This is the circle centered at l i with radius 1 It is shown below 0 r 5 a Write lz 4iz4i 10 as Iz 4illz 4il 10 to see that this is the locus of all points 2 such that the sum of the distances from 2 to 4139 and 4i is a constant Such a curve is an ellipse with foci i4i b Write lz 1lzi as lz 1z il to see that this is the locus of all points z such that the distance from39 z to 1 is always the same as the distance to i The curve is then the perpendicular bisector of the line segment from 1 to i SECTIONS 1 a 21 3iz3i b iz39z iz c Wm22 i24 4ii24 4i 134i d 225 i22 5 2 2z 5V21 5 2z5 2 a Rewrite Rez i 2 as Rex i y 1 2 or x 2 This is the vertical line through the point z 2 shown below b Rewrite 22 ii 4 as 2 z 4 or z 2 This is the circle centered at 3 with radius 2 shown below 2 i2 1 0 Write zl xl in and 22 x2 iy2 Then 21 22 x1 in x2 in xx x2iy1 yz x1x2iy1 y2x1 iyi x2 i 2 21 22 and 39z lz Z x1 i 1x2 135 x1x2 Y13 2 10le 35135 xi7 2 quot 3 13 2 i 1x2 x1y2 x1 iylx2 139 2122 a z122233 zlzlz3 1 1 b 2zzzz2z z zzzzzz3222zz24 2233 zzza 22 la 7 z Izl lzl b 1 1 1 39 2223 Izzzgl Izzllz3l In this problem we shall use the inequalities see Sec 4 RezlSzl and lz z2 z3 S Izllzz 23 Specifically when IzlS 1 Re2zz3ltI2zz3lszlzl Iz3l 2lzllzl3s 211 4 10 First write Z4 43 3 z2 1z2 3 Then observe that when z 2 and Ir2 112Iz2II1IIzIZ 1I4 1I 3 Iz2 3I2zzl I3IIzI2 3I4 3I1 Thus when z 2 Iz 422 3ttz2 1zz 3 2 31 3 Consequently when 2 lies on the circle Izl 2 11 a b 12 a Ill1 5 Iz44z23l lz4 4z23l l 3 Prove that z is real gt 2 2 4 Suppose that 2 2 so that x iyxiy This means that i2y0 or y0 Thus zxi0xor zisreal gt Suppose that z is real so that z xi0 Then 2 x i0xi0z Prove that z is either real or pure imaginary ltgt 22 zz lt Suppose that 2392 zz Then xiy2 xiy2 or i4xy 0 But this can be only if either x 0 or y O or possibly x y 0 Thus z is either real or pure imaginary gt Suppose that z is either real or pure imaginary If z is real so that z x then 22 x2 zz If 2 is pure imaginary so that z iy then 22 iy2 iy2 22 We shall use mathematical induction to show that ahmaimi n 23 This is known when n 2 Sec 5 Assuming now that it is true when n m we may write zr Z2 zm zmr2122 zzmr z1z2 quot39Zmzm1 21 Zz39 zmzm1 51 224quot quotan zml39 A b In the same way we can show that ztzzmzn zizz39 39zn n23 This is true when n 2 Sec 5 Assuming that it is true when n m we write 2122 39quotzmzmu 2122 zmzm1 2122 quotzm EM 2122 2quot zm 2122 2mm zz enable us to write lz zol R as 14 The identities Sec 5 zz zl2 and Rez 2 20550 R2 zz zzoEz Tzozo R2 Izlz 2RezZo Izol2 R2 15 Since x 5123 and y 5 24 the hyperbola x2 y2 1 can be written in the following l ways are Er ii 2 2i z22z f22 z2 2zz72 1 4 4 222 222 1 4 9 2 SECTION 7 1 a Since i 2 arg22i argt arg 2 1 i 15 37 57 one value of arg IS or Consequently the pnnCIpal value is 2 21 2 4 4 57 37 27 or 4 4 5 b Since age5 i 6arglt i one value of arg i6 is 676 or 71 So the principal value is 39 27 or 7 The solution 9 7 of the equation lei 1 2 in the interval 0 S 0 lt 271 is geometrically evident if we recall that equot9 lies on the circle zl and that lew II is the distance between the points em and 1 See the gure below We know from de Moivre39s formula that cos 6 i sin 63 c0830 i sin39 cos3 9 3cos2 6i sin 9 3cos 0i sin 62 i sin 93 cos 39 i sin 39 That is cos3 9 3cos 93in2 15liBCos2 6sin 0 sin3 6 c0536 isin36 By equating real parts and then imaginary parts here we arrive at the desired trigonometric identities a c0536 cos3 0 3cos sin2 9 b sin 39 3cos2 Gsin 0 sin3 9 Here 2 re is any nonzero complex number and n a negative integer n 12 Also m nl2 By writing 1 ZM 1rmeim9 1quotTex m0 zlm lea91 quot ei m9 1eim9 r r39quot r and we see that zquot39391z39lquot39 Thus the definition 2quot z39l can also be written as Zn zmyl 10 10 First of all given two nonzero complex numbers zl and z suppose that there are complex numbers c1 and c2 such that z1 clc2 and 22 c152 Since z1lc1lczl and z2lclllc2llclllc2 it follows that lzllz2 Suppose on the other hand that we know only that Izlllz2L We may write ZirreXPi91 and z2r1expi92 If we introduce the numbers c1 r1 expi 6 92 and c2 expi 9 62 2 2 we find that clcz r1 expi 6 Z 62 expi r1 expi91 21 and c r1 expi 6 Z 62 expi 6 g 02 j rl exp 62 zz That is zlclc2 and zzc152 IfS1zz2zquotthen S zS 1zz2zquotzz2 z3zquot 1 2 n1 Hence S 1 z provided z 1 That is z 1 n1 I Za l 1zz2zquot putting z e 0 lt 9 lt 27 in this identity we have I inl9 1e 9 e3929equotquot 1 e 4 1e19 11 Now the real part of the lefthand side here is evidently 1 c056 cosZGcosn6 and to find the real part of the righthand side we write that side in the form ex 49 ex 43 1 expinMI P 2 P 2 p 2 1 expae exp i exp i expi which becomes 9 6 2n10 2n19 COSiISIHECOS ISIn i 2isin 2 or sin2 sin 2n 10 iltos2 cos 2n 16 2 2 2 2 23m 2 The real part of this is clearly 2n 10 1 Sin 2 6 25m and we arrive at Lagrange s trigonometric identity Sin 2n 19 1 2 1cosGcos29mcosn9 T 0lt 9lt27r 2sin 2 12 SECTION 9 1 a 17 Since 2i 2expi 2kn k Oili2 the desired roots are 2012 E expi kn k 01 That is c em4 cos isin iL o 4 4 5 1 1 and c1 Hem e 00 1 i co being the principal root These are sketched below y 15 x Observe that 1 5139 2expi g 2kn k 0iliz Hence 1 43012 Viexpi 67 kn k 01 The principal root is 00 View VZ cosg ising and the other root is cl eiz6eiz C0 i These roots are shown below 2 13 a Since 16 16expi7z 2k7z k 0ili2 the needed roots are 16u4 zcxpik 2 l k 0123 The principal root is C 2ei7r42cos isin 21L o 4 4 2 5 50 The other three roots are c1 2equotquot 4equotquot 2 coi J5 1 ii 45 1 i CZ zein39l4ein c0 1i and c3 2equot equot quot2 co i V5 1 i i 5 1 i The four roots are shown below b First write 8 843i 16expi 33 2mg k 01112 Then 8 Nit Zexpi k 0123 2 The principal root is c 2e quotquot2cos is39n 2 6 1 6 2 2 5 quot 14 The others are cl 2equot quot5e quot2 coi 1 5139 c2 2ei 6ei7r co c3 2equotquotquot e quot 2 co i 1 i These roots are all shown below y 3 a By writing 1lexpi2t 2k7t k 0ili2 we see that 1113 expi7 k 012 3 3 The principal root is co em 3 cos isinE 1J i 3 3 2 The other two roots are and 395 3 71 71 1 i 2er I e121re mlscoslsm 3 3 2 All three roots are shown below y 15 b Since 8 8expi0 2km k 0ili2 the desired roots of 8 are 8 quot 5 apple k o123 45 the principal one being c0 39239 The others are cl fem3 cos ising gi I VEgt 4713 i1 1quot i cz wfe e c0s3 lsm3 1 452 2 l T cg J ei 9 in i 1 and 113 cs ei2 3ei cz V I All six roots are shown below The three cube roots of the number 20 4 445i 8expi are evidently z0113 Zexpi k 012 In particular co Zexpi 45 1 i 16 With the aid of the number wsl i we obtain the other two roots 1 339 3 1 3 1 01 Cow3 1l 3ft 4 1 1 139 1 339 31 3139 czzcow cow3w3 l l 5 a Let a denote any xed real number In order to find the two square roots of ai in exponential form we wnte Aail a2 1 and aArgai Since aiAexpia2k7r k Oi1i2 we see that ai 2 wZexpikn k 01 That is the desired square roots are ew2 and ew28m Z39eia239 b Since a H lies above the real axis we know that O lt a lt 71 Thus 0 lt g5lt 325 and this tells us that cost gt 0 and sin gt 0 Since cos a i it follows that cosg 1cosaL 1 VAa 2 2 El A and sing A a 2 V 2 El A EJZ39 Consequently Aa A a i add4 2 9i A v4 V e 0082 181112 1amp4 i 13wmiwZ Z 17 6 The four roots of the equation z4 4 0 are the four fourth roots of the number 4 To nd those roots we write 4 4expi7z39 2k71 k Ot1i2 Then 414 z expli17 ein4eikmz k LL23 To be speci c c0 wfe39m w cosisin 45712 i J 1i c1 coe m lii w1i c2 coe 1 il 1 i c3 coemquot2 1 i i 1 i This enables us to write Z4 4 Z Coxz cizquotczz cs Z 012 390239Z CoZ 63 z1iz1i39z 1 iZ1i zl2lz l21 z2 22 2X22 22 2 7 Let c be any nth root of unity other than unity itself With the aid of the identity see Exercise 10 Sec 7 1zzZzquot 1z 5 1 z wefindthat 1cc2cquot391 1 c l 10 l c 1 0 9 Observe first that 1 I Zlm1 exp fieITexPL Iizk m exp 119 exp l 2nk 18 and z1lm Iexpl62k7r 1 r m V m where k 012 m 1 Since the set i2k7t m exp is the same as the set exp 12k7t m but in reverse order we find that zquotquot 1 z 11M SECTION 10 1 6X i2k7 p m k 012m 1 k012m 1 a Write lz 2 1151 as lz 2 iSl to see that this is the set of points inside and on the circle centered at the point 2 i with radius 1 It is not a domain b Write l223lgt 4 as re gt 2 to see that the set in question consists of all points exterior to the circle with center at 32 and radius 2 It is a domain 19 0 Write Imz gt1 as y gt1 to see that this is the half plane consisting of all points lying above the horizontal line y 1 It is a domain 0 x d The set Imz 1 is simply the horizontal line y 1 It is not a domain y1 e The set 0 S argz S E z 0 is indicated below It is not a domain f The set lz 42z can be written in the form x 42 y2 2 x2 y which reduces to x S 2 This set which is indicated below is not a domain The set is also geometrically evident since it consists of all points 2 such that the distance between 2 and 4 is greater than or equal to the distance between z and the origin 20 4 a The closure of the set 7r lt argz lt 7 z ab 0 is the entire plane b We first write the set lRezlltlzl as xlltqlx2 yz or x2 ltx2y2 But this last inequality is the same as y2 gt 0 or I ygt 0 Hence the closure of the set lRe zlltlzl is the entireplane c Since l zL2 2 Lyz the set Relsl can be written as z x z Sl or zzzlzl xy z 2 xy 2 x2 2x y2 2 0 Finally by completing the square we arrive at the inequality x l2 y2 2 12 which describes the circle together with its exterior that is centered at 2 1 with radius 1 The closure of this set is itself 21 d Since z2 1 1302 362 y2 i2xy the set Rez2 gt O can be written as y2 lt x2 or yltlx The closure of this set consists of the lines y ix together with the shaded region shown below 5 The set S consists of all points z such that lzllt l or lz 2lt1 as shown below Since every polygonal line joining Z1 and z2 must contain at least one point that is not in S it is clear that S 18 not connected 8 We are given that a set S contains each of its accumulation points The problem here is to show that S must be closed We do this by contradiction We let 20 be a boundary point of S and suppose that it is not a point in S The fact that zo is a boundary point means that every neighborhood of z0 contains at least one point in S and since 20 is not in S we see that every deleted neighborhood of S must contain at least one point in S Thus 20 is an accumulation point of S and it follows that 20 is a point in S But this contradicts the fact that 20 is not in S We may conclude then that each boundary point zo must be in S That is S is closed 22 Chapter 2 SECTION 11 1 1 a The function fz 1 18 de ned everywhere in the finite plane except at the z2 points z ii where 22 1 O 1 b The function I f z IS defined throughout the entire nite plane except for the point z 0 c The function fz is defined everywhere in the nite plane except for the imaginary axis This is because the equation z E 0 is the same as x 0 d The function fz is defined everywhere in the finite plane except on the 1 1 Izl2 circle Izl 1 where l lzzl2 0 22 zZ 3 Usmg x and y wnte 2 21 fz x2 y2 2yi2x2xy 2 2 z zziizzizz2ZXZ Z 4 4 2 2 2 2 z z z z 2 quot 39l 2 39 2 2 12 2 2 z21z SECTION17 5 Consider the function 2 2 fzf z 0 z x ly where z x iy Observe that if z x0 then fz 1 x iO and if z OJ 0iy 2 fZ 0iy 439 10 11 23 But if z xx xix 2 1i 2 fz J 1 This shows that fz has value 1 at all nonzero points on the real and imaginary axes but value 1 at all nonzero points on the line y x Thus the limit of fz as 2 tends to 0 cannot exist 2 4 z 4 we use statement 2 Sec 16 and write z 1 1 2 42 4 lim a To show that lim 20 I 2 0 1 02 4 z b To establish the limit lirrli 1D3 co we refer to statement 1 Sec 16 and write z z l 3 931z13 1 3z 1 039 2 c To verify that lim 2 11 oo we apply statement 3 Sec 16 and write 2 z 1 E 1 z z2 2quot 2 J H z z In this problem we consider the function azb ad beam czd Tz a Suppose that c 0 Statement 3 Sec 16 tells us that lim Tz oo since 2 lim L 11 E O z 0 Tlz HO abz a 24 17 Suppose that c 0 Statement 2 Sec 16 reveals that lim Tz 9 since zw 5 lim 11 lim bz 3 0 z 40 cdz c Also we know from statement 1 Sec 16 that linil Tz 00 since z c Hm 1m czd 0 z dc Tz z dc az b SECTION 19 1 a If fz 3z2 2z4 then f 2 d39322 2z4 3122 2 dzi4 32z 210 62 2 dz dz dz dz 39 b If fz1 4z23 then f z 31 4z 2 in 48 31 4z22 8z 24z1 4f l 1 c Iffz2zzl me 3 then d d fz 22 1jdz z 1 Z 12z2z 1 2z 11z 12 3 W D2 2z 12 225 12 39 1 Z2 4 22 z 0 then d If fZ 722 zzy 221z2342 Z 1 22 21 253322 1 Z4 23 39 25 3 If fziz z 0 then Az A AZ 39 w fz fz 2 z z z Hence Aw 1 z 11m lim f AHOAZ AZ zAzz 22 4 We are given that f 20 gzo 0 and that f zo and g39zo exist where g z0 0 According to the de nition of derivative z uo z zo 24 z 20 Similarly SW lim 3z g20 lim 32 1920 z zo z no z 20 Thus lim L lim fZZ 20 1132 m z z 10 We 80 Mo 82 z Zo 82 z Zn 20 SECTION 22 1 a fzZx iy So u x v y b c d Inasmuch as ux vy gt1 1 the CauchyRiemann equations are not satisfied anywhere fzzZ39xiy x iy0i2y So u 0 v 2y Since u v O 2 the CauchyRiemann equations are not satis ed anywhere fz 2x ixy2 Here u 2 v xyz u vy gt22xygtxyl u v gt0y2 gty0 Substituting y 0 into xy 1 we have 0 1 Thus the CauchyRiemann equations do not hold anywhere fzequotequot39y e cosy isinye cosy ie siny So ue cosy v equotsiny u vy gte cosy e cosygt 2equot cosy0gt cosy0 Thus y12 n77 n0i112 uy v gt e siny e siny 9 2e siny 0 siny 0 Hence yn7r n0ili2 Since these are two different sets of values of y the CauchyRiemann equations cannot be satis ed anywhere 26 1 1 Z Z x y 3 a Z 39 1 SO u and v x2y2 x2y2 Since 2 2 y x ny 2 2 u v and u v x 0 1 x2y22 Y y x2y22 I y f z exists when 2 0 Moreover when 2 96 0 2 2 2 y x 2xy x l2xyy f z quotx lvx x2y22 lx2y22 x2y22 3 iygt2 i y22 zzf 39 may zl39 b fzx2iy2 Hence ux2andvy2 Now xvygt2x2ygtyx and uy vxgt00 So f z exists only when y x and we nd that f x ix uxx ivxxx 2x i0 2x c fz zImz xiyy xyiy2 Here u xy and vy2 We observe that u vygty2ygty0 and uyvxgtx0 Hence f z exists only when 2 0 In fact f 0 u00 ivx00 0 i0 O 4 a fzcos40i sin40 ziO Since 2 Lw u V ru 139 ri4cos40 v9 and us 47sin46 rv 5 27 f is analytic in its domain of definition Furthermore f 39z e39i9u iv ijcosw i sin 40 r 3 e quot cos 46 i sin 40 ise39 e r r 4 4 i rseise rei95 Z5 39 b fzw7ei9 2w7cos ixsing rgt0alt0lta27r Since W N v I a I 9 ru cos v9 and u 9 sm rv 2 2 2 2 r f is analytic in its domain of definition Moreover f 39z equotau W Z I cosg i sin g 2 1 i9 9 9 1 49 fall Z e 2 lmz ZWe e 1 1 21ei02 2fz39 c f z e 9 cosln r ie e sinln r r gt 00 lt 6 lt 27 Since v J W J u v ru e399 sinln r v9 and u e399 cosln r rv f is analytic in its domain of de nition Also f 39z e39i9u iv Kiel e 9 sinln r 1 e cosln 0 r r E z Liae4 cosln r ie39e sinln r 1 re When fz x3 i1y3 we have u x3 and v 1 y3 Observe that uxvygt3x2 31 y2gtx21 y20 and uy vxgt00 28 7 Evidently then the CauchyRiemann equations are satis ed only when x O and y 1 That is they hold only when 2 i Hence the expression f z u iv 32 i0 3x2 is valid only when z i in which case we see that f i 0 Here u and v denote the real and imaginary components of the function f de ned by means of the equations 2 fz 2 when z 0 z 0 when z0 Now 1 fZ x3 3xy2i y33x2y x2y2 x2y2 H 9 when z at O and the following calculations show that u 00 v 00 and uy 00 v00 u0 Ax0 u00 lim 9 u lim 11 x lit0 Ax Air 0 Ax u 00 MW 11m0 0 y Ay Ay 0 v 0015m Ax Ov 9 mug0 Ax ao Ax mo Ax v 00 km quot OAy lim AV 1 y Ayn0 quo Equations 2 Sec 22 are uxcos9uysin9 u ursm 9 uyrcos 6 us 29 Solving these simultaneous linear equations for ux and u we find that sine 0059 and u usm6u9 ur ucos9 u9 Likewise sine cos0 vvcos9 v9 and vyvsm0v9 r Assume now that the CauchyRiemann equations in polar form ru v9 ua rv are satisfied at zo It follows that cos9 sine cosG vsm6vsm0v9 vy r ve ur ucosG u9 0080 sine sine uyurs1n0u9 v 9 vcos9 vrcosa va vx r r r 9 a Write f z ur 0 ivr 9 Then recall the polar form ru v9 14 rv of the CauchyRiemann equations which enables us to rewrite the expression Sec 22 f 39zo e39w ur iv for the derivative of f at a point z r0 90 in the following way i9 1 i J 1 i u 1v u 1v fzo e rve r o rem a a Zo a 9 30 b Consider now the function 1 1 1 1 H fz w e cos9 zsme ls i Z re r r With ur9c se and map 993 r the final expression for f zo in part a tells us that i Z r r z l 59 l1i z r 2 re 252 r when 2 6 0 10 a We consider a function F x y where x d zz39 Formal application of the chain rule for multivariable functions yields 8F t 1 8F 113F 195 EE axazayaz ax 2 3 3927 EEHay39 b New define the operator 1 1 8 8 32 2 Bx By suggested by part a and formally apply it to a function f z ux y ivx y ilifiiliii 32 2 ax ay zax 29y 50 ivxuy iv21 x Vyivx uy39 If the CauchyRiemann equations u vy uy vx are satisfied this tells us that 8f820 3 1 SECTION 24 l a fz 3x y i3y x is entire since T l ux3vy and uy 1 v b f z sinxcosh y icosxsinh y is entire since d ha ll V uxcosxcoshyvy and uysmxsmhyvx e cosx mx C fz e smx l e i l e cosx1sent1re smce V ux equot cosx vy and uy equot sinx v d f z z2 2 quot2 iy is entire since it is the product of the entire functions gz z2 2 and hz e e equotcos y i sin y equot cos y i equot sin y W h T J l The function g is entire since it is a polynomial and h is entire since x x u e cosyvy and u e smy vs 2 a f z a iy is nowhere analytic since uxvygtyl and uy vxgtx0 which means that the CauchyRiemann equations hold only at the point z 01 i y it y y y 39 c c fz e e e cosxzsmx e cpsxze smx IS nowhere analytic sm e V u v 5 e sinxe sinxgt 2e sinx0gtsinx0 and uy v gt e cosx e cosx gt 2e cosx 0 3 cosx 0 More precisely the roots of the equation sinx O are rm n 0t1i2 and cosmr 1quot 0 Consequently the CauchyRiemann equations are not satis ed anywhere 7 a Suppose that a function f z ux y ivx y is analytic and realvalued in a domain D Since f z is realvalued it has the form f z ux y 0 The CauchyRiemann equations u v u vx thus become u 0 u 0 and this means that ux y a where a is a real constant See the proof of the theorem in Sec 23 Evidently then f z a That is f is constant in D 32 b Suppose that a function f is analytic in a domain D and that its modulus I fz is constant there Write Ifzlc where c is a real constant If C O we see that f z 0 throughout D If on the other hand c at 0 write f am 2 or CZ fZ Since f z is analytic and never zero in D the conjugate 7amp3 must be analytic in D Example 3 in Sec 24 then tells us that f z must be constant in D SECTION 25 it 3 1 a It is straightforward to show that up uyy 0 when ux y 2xl y To find a harmonic conjugate vx y we start with u x y 2 2y Now u vygtvy 2 2ygtvxy2yy2 x Then it v 9 2x x gt x 2x 9 x x2 c Consequently vxy 2y y2 x739 c x2 y2 2yc b It is straightforward to show that u u 0 when ux y 2x x3 3xy2 To find a harmonic conjugate vx y we start with uquot x y 2 3x2 3y2 Now u vy gt vy 2 3x2 3y2 5 vxy 2y 3x2yy3 x Then u vx gt 6xy 6xy x 9 x 0 gt x c Consequently vxy 2y 25ny y3 c c It is straightforward to show that un uyy 0 when ux y sinhxsin y To nd a harmonic conjugate vx y we start with ax x y coshxsin y Now u v gt v coshxsiny gt vxy coshxcosy x Then uy vx gt sinhxcos y sinhxcos y 39x gt x 0 gt x c Consequently vxy coshxcosy c 33 d It is straightforward to show that u 141y 0 when uxy i 2 To find a y x2 ny harmonic conjugate vx y we start With ur x y 7 2 W Now x y 2xy x quot 3 V W 21 M Then x2 yz x2 y2 394 Vx gt xgt x0 xc y x2 y22 x2 y22 Consequently X V y x2 y c Suppose that v and V are harmonic conjugates of u in a domain D This means that ur vy uy vx and u Vy uy Vx If wv V then wxvx Vx uyuy0 and wyv Vyu ux0 Hence wx y c where c is a real constant compare the proof of the theorem in Sec 23 That is vx Vx y c Suppose that u and v are harmonic conjugates of each other in a domain D Then ux v u vx and Vx uy vy ux It follows readin from these equations that ux0 uy0 and vx0 vy0 Consequently ux y and vx y must be constant throughout D compare the proof of the theorem in Sec 23 The CauchyRiemann equations in polar coordinates are ruv9 and u9 rv Now rurv9gtrunurver 34 and us rvr gt rvr9 rzuquot ru u99 We rvm and since v9 9 we have rzuquot ru um 0 which is the polar form of Laplace39s equation To show that v satisfies the same equation we observe that 1 l u rv gtV u V u u 9 r r r 9 rr r2 9 r 9r and rurv9gtv99ru9 Since 149 143 then rzvrrrvv99 ue ru9 u9ru 0 Ifur9lnr then 2 2 1 1 runruuaar r 2 r O0 This tells us that the function u lnr is harmonic in the domain r gt 0 O lt 6 lt 271 Now it follows from the CauchyRiemann equation ru v9 and the derivative u 1 that v9 1 r thus vr9 9 r where r is at present an arbitrary differentiable function of r The other CauchyRiemann equation us rv then becomes O r r That is r0 and we see that r c where c is an arbitrary real constant Hence vr6 9 c is a harmonic conjugate of ur6 1n r 35 Chapter 3 SECTION 28 1 a exp2 i 3m ezexpi3m39 e2 since expi37ri 1 2m b exp 4 ex l6x 42cos isinz p 2 p 4 4 4 JZliLE1i 39239 fi 2 39 c expz xi expzexp m39 expz since exp ni 1 3 First write exp239 expx iy e e39 equot cos y iequot sin y where z x iy This tells us that expZ ux y ivx y where ux y e cosy and vx y e sin y Suppose that the CauchyRiemann equations ux v and uy v are satis ed at some point z x iy It is easy to see that for the functions u and v here these equations become cos y O and sin y 0 But there is no value of y satisfying this pair of equations We may conclude that since the CauchyRiemann equations fail to be satisfied anywhere the function expZ is not analytic anywhere 4 The function expzz is entire since it is a composition of the entire functions 22 and expz and the chain rule for derivatives tells us that Zexpzz expzz1dz2 22 expzz Alternatively one can show that expz2 is entire by writing expzz expx iy2 expx2 y2expi2xy expx2 y2cos2xy i expx2 y2 sin2xy ll V and using the CauchyRiemann equations To be speci c ux 2x expx2 y2cos2xy 2 y expc2 y2sin2xy vy and uy 2 y expx2 y2cos2xy 2x expx2 y2sin2xy vx Furthermore dexpzz u iv 2x iyexpx2 y2cos2xy iexpx2 y2sin2xy dz 2z expz2 We rst write Iexp2z i Iexp2x i2y 1 ez and lexp z2 l lexp 2xy ix2 y2 e39zquot Then since lexp2z i expiz2 l S exp22 i l lexpiz2 l it follows that lexp2z i expizz S equot e39z First write lexpltzz lexpux iygt21 Iexpx2 yz i2xy expx2 y and expz2 expx2 y2 Since c2 y2 S x2 yz it is clear that expx2 yz S expx2 yz Hence it follows from the above that lexpltz2 s expazlz To prove that Iexp 2z lt 1ltgt Rez gt 0 write lam22 lexplt 2x i2ygtl explt 2x It is then clear that the statement to be proved is the same as exp 2x lt 1ltgt x gt 0 which is obvious from the graph of the exponential function in calculus 8 a Write e1 2 as exeiy 2e This tells us that e 2 and y7r2mt That is xln2 and y 2n17 Hence zln22n11ri b Write e2 l i as e equot Ze W from which we see that e 2 and y 2mr That is 1 x an and y2n Consequently zln22n 17ti Write exp2z 1 1 as ah1e 1equot and note how it follows that 0 e2quot1 1 and 2y O2n7r Evidently then 1 d it x an y n and this means that zlmri 2 This problem is actually to nd all roots of the equation expiz expi2 n Oili2 n 0i39lt2 n Oi39li2 n 0ili2 n 0ili2 n 0ili2 n Oi1i2 n 0ili2 n 0ili2 37 38 10 12 13 To do this set z x iy and rewrite the equation as yix ee y ix e e Now according to the statement in italics at the beginning of Sec8 in the text equote and xx2mt where n may have any one of the values n 0ili2 Thus y0 and xn7r n0il2 The roots of the original equation are therefore zn7r n0i1i2 1 Suppose that e2 is real Since ez equot cos yiequot sin y this means that equot sin y 0 Moreover since e r is never zero sin y 0 Consequently y nil n Oili2 that is Imz n71 n 0ili2 b On the other hand suppose that e is pure imaginary It follows that cosy 0 or that y mr n 035112 That is Imz gmr n 0i12 We start by writing x iy x y l x2y2 x2y2 x2y2 Because Ree equot cos y it follows that R1quot x os y Lcos L Ce exPx2y2C x2y2 exP x2y2 x2y2 Since equot is analytic in every domain that does not contain the origin Theorem 1 in Sec 25 ensures that Ree is harmonic in such a domain If f z uxy ivx y is analytic in some domain D then em equotquot cos vx y iequotquotquot sin vx y Since em is a composition of functions that are analytic in D it follows from Theorem 1 in Sec 25 that its component functions U x y equotquotquot cosvx y Vx y e quot sin vxy 39 are harmonic in D Moreover by Theorem 2 in Sec 25 Vx y is a harmonic conjugate of U x y 14 The problem here is to establish the identity exp 2quot expnz n 0i1i2 a To show that it is true when n0l2 we use mathematical induction It is obviously true when n 0 Suppose that it is true when n m where m is any nonnegative integer Then 614quot11 exp Zquotl exp 2 expmzexp z expmz z expm 1z b Suppose now that n is a negative integer n l 2 and write m n 12 In view of part a ex znlm 1 1 1 ex n p expz expzquot expmz exp nz p z SECTION 30 1 a b 2 a b c 3 a Log ei lnI eiiArgei lne i 1 125i 717 l Log1 ilnl1 iliArgli ln i 21 1n2 4 loge lnei02mr 12nm39 n 0ili2 logi 1n1i 252mc 2n7ri n 0i1i2 log 1J i ln2i2 3 2mt ln22nn i n 0ili2 Observe that Log1 i2 Log2i 1n2 325i and 75 39 2Log1l 2lm 2zz ln2 2 1 Thus Log1 i 2Log1 i 40 b On the other hand Log 1i2 Log2i ln2 i and 2Log li 21m i ln23 2 i Hence Log 1i2 2Log 1 i 4 a Consider the branch 7t 97 logzlnr10 rgt0ZltGlt Z Since logiz log l ln1i7t m and Zlogi 2lnli 72 7ri we nd that logiz Zlogi when this branch of logz is taken b Now consider the branch logzlnri0 rgt0 lt6lt11 n 4 4 Here logiz logl ln1i7r m and Zlogi 2ln1i57 571i Hence for this particular branch logiz 2 Zlogi 5 a The two values of i 2 are equotquot 4 and emquot Observe that am 7F 1 loge lnllz2nn2nzm n0i1i2 and logei5 4ln1i Znn 2n 1 7l39i n 0112 Combining these two sets of values we nd that logim n n 0ili2 7 To solve the equation logz MI 2 write explogz expi7t 2 or z e b 41 On the other hand 1 1 7r 1 l 1 1 2 2 0g in 2 nnj n4m 12 Thus the set of values of logz n 012 is the same as the set of values of logi and we may write logi 2 llogi 2 Note that logi2 log1 1n17r 2mri 2n 1m39 n Oili2 but that 210gi 2ln1i2mrl 4n1m39 n 01r2 Evidently then the set of values of logi2 is not the same as the set of values of Zlogi That is logi2 210gi int2 1 10 Since lnx2 y is the real component of any analytic branch of Zlogz it is harmonic in every domain that does not contain the origin This can be veri ed directly by writing ux y lnx2 yz and showing that u x y u x y 0 t ESECTION 31 1 Suppose that Re zl gt 0 and Re z2 gt 0 Then zl r1 exp i9l and 22 r2 exp i82 where lt91lt 75 and lt2lt 2 2 2 2 42 The fact that 71 lt 91 92 lt 11 enables us to write Logzlz2 Logrlr2exp il 92 lnrr2 i1 92 1n rl 91 1n r2 i632 Logrl exp i631 Logr2 exp 13902 Logzl Logzz 3 We are asked to show in two different ways that 10g 10321 10gzg 21 0 22 0 52 a One way is to refer to the relation arg 1 argzl argz2 in Sec 7 and write 32 log z ln 22 Z b Another way is to first show that 10 logz z 0 To do this we write 2 rei z iarg2 1lnlleiargzl lnlz2liargzzlogzllogz2 2 9 and then logeIogGe39iquot1n1i a 2717 ln r i0 mm logz Z where n 0ili2 This enables us to use the relation 1033132 10g Zr log Zz and write zz zz Z2 43 5 The problem here is to verify that ln 1 2 exp zlogz n 1 2 given that it is valid when n 12 To do this we put m n where n is a negative integer Then since m is a positive integer we may use the relations zquot1 1 z and 1 e equot to write 1 z1 zlm 1 expi 10g z m expfi 103 exp 1 logz expllogz m m n SECTION 32 1 In each part below n 01i2 a 1 i expilog1i expi lnwf Mull exp ln 2 Zn exp 21m expGan nce n takes on all integral values the term 2n7r here can be replaced by 2mt us 39 1 1 exp 4 2nnexp2 m2 l17 expi logl expln 1 13901 2n7t exp2n 2 a PV iquot expiLogi expiln1 emfg 0 PV39 EH 39 50 exp3mLog I 430 39exp3m39lne exp27r2expi37t exp2752 0 RV 1 i exp4iLog1 1 exp4ion ezeim equotlC S4ln 5 i sin41n I 1 equotcos2 ln 2 isin21n 2 Since 1 i Zez m we may write 1 5032 expglog l 50 expln 2 2n7r expln2m 3n 17ri 2V39239exp3n 1139 where n Oili2 Observe that if n is even then 3n 1 is odd and so exp3n17ri 1 On the other hand if n is odd 3n1 is even and this means that exp3n 17ri1 So only two distinct values of 1 5032 arise Speci cally 1 Vii 1245 We consider here any nonzero complex number 20 in the exponential form z0 r0 exp iGO 9 where 7r lt 90 S 7 According to Sec 8 the pnncrpal value of 2 quot IS r0 expt l and n according to Sec 32 that value is 1 1 6 9 Lo ex 1n 9 1n 39 39 expn g2 Pn r0 1 0 exp roexpz n ro expl n These two expressions are evidently the same Observe that when cabi is any fixed complex number where c 0ili2 the power i c can be written as i expclog i expa biln1 burn 7t 7 exp b3 2n la 2 2n7t n 0ili2 Thus lic exp b2mr n 012 and it is clear that li l is multiple valued unless b 0 or c is real Note that the restriction c at 0i1i2 ensures that ic is multiplevalued even when b 0 45 SECTION 33 1 The desired derivatives can be found by writing isinzi ell ed i 1eiz ieiz dz dz 2i 2i dz dz iz 21 iie ziequot ez cosz and d Eu elz 1 d eiz d e iz dz dz 2 2 dz dz 1 39 i ei e lls1nz 2 1 21 2 From the expressions ell 1 eiz eiz smz and cosz we see that 12 iz iz iz e e e coszzs1nz e 3 Equation 4 Sec 33 is 23in 21 cos z2 sinzl zz sinzl 22 Interchanging 21 and 22 here and using the fact that sin z is an odd function we have 39 2cos zl sin 22 sinz1 22 sinzl z2 Addition of corresponding sides of these two equations now yields 2sin z1 cosz coszl sin 22 25inzI z sinzl zz sin zl cos 22 cos zI sin zz 4 Differentiating each side of equation 5 Sec 33 with respect to 21 we have coszl zz cos 21 cos z2 sin zl sin 22 46 7 10 11 a From the identity sin2 2 cos2 z 1 we have sinzz coszz 1 coszz coszz coszz or 1tanzzseczz b Also 2 2 s1n 2 cos 2 1 T z 2 2 or lcotzzcsczz sm 2 sm 2 sm z From the expression sinz sinxcosh y i cosxsinh y we find that lsin zl2 sin2 xcosh2 y cos2 xsinh2 y sin2 xl sth y 1 sin2 Jasmin2 y sin2 x sinh2 y The expression cosz cosxcosh y i sinxsinh y on the other hand tells us that lcoszl2 cos2 Jccosh2 y sin2 xsinh2 y cos2 xl sinhz y 1 cos2 x sth y 0032 x sinh2 y Since sinh2 y is never negative it follows from Exercise 9 that a Isin 22 2 sin2 x or lsin 2 2 Isian and that b Icoszl2 2 cos2 x or Icos 2 2 Icos xl In this problem we shall use the identities lsin 22 sin2 x sinh2 y lcos zl2 cos2 x sinh2 y 47 a Observe that sinh2 y sin zl2 sin2 x s Isin zl2 and Isin zl2 sin2 x cosh2 y 1 cosh2 y 1 sin2 x cosh2 y cos2 x S cosh2 y Thus sinh2 y S Isin zl2 S cosh2 y or lsinh ylSlsinzlS cosh y b On the other hand sinh2 y Icos zl2 cos2 x S Icos zl2 and Icos zl2 cos2 x cosh2 y 1 cosh2 y 1 cos2 x cosh2 y sin2 x S cosh2 y Hence sinh2 y lt Icos zl2 S cosh2 y or Isinh yl S lcoszls cosh y 13 By writing f z sin sinx iy sinxcosh y i cosxsinh y we have f z uxy ivxy where ux y sinxcosh y and vx y cosxsinh y If the CauchyRiemann equations u v uy vr are to hold it is easy to see that cosxcosh y 0 and sinxsinh y 0 Since cosh y is never zero it follows from the rst of these equations that cosx O that is x gmr n O i 1iZ Furthermore since sinx is nonzero for each of these values of x the second equation tells us that sinhy 0 or y 0 Thus the CauchyRiemann equations hold only at the points z 725mr n0i1t2 Evidently then there is no neighborhood of any point throughout which f is analytic and we may conclude that sin is not analytic anywhere The function f z cos cosx iy cosxcosh y i sinxsinh y can be written as f z m0 ivxy where ux y cosxcosh y and vx y sinxsinh y 48 If the CauchyRiemann equations ur V uy v hold then sinxcoshy 0 and cosxsinhy 0 The first of these equations tells us that sinx0 or xmrn0ilt2 Since Cosmn o it follows that sinhy0 or y0 Consequently the CauchyRiemann equations hold only when zmr n0ili2 So there is no neighborhood throughout which f is analytic and this means that cos39z is nowhere analytic 16 a I Use expression 12 Sec 33 to write 53 W cosycoshx isinysinhx and cosi239 cos y ix cos ycoshx i sin ysinh x This shows that cosiz cosi39z39 for all z Use expression 11 Sec 33 to write siniz siny ix sin ycoshx i cos ysinhx and siniz siny ix sin ycosh x i cos ysinh x Evidently then the equation siniz siniz is equivalent to the pair of equations sinycoshx 0 cosysinhx 0 Since coshx is never zero the first of these equations tells us that siny 0 Consequently y rm n 0l1i2 Since cosmr 1quot 6 O the second equation tells us that sinhx O or that x 0 So we may conclude that siniz siniZ if and only if z 0 imt nm39 n Oi1i2 17 Rewriting the equation sinz cosh4 as sinxcosh y i cos xsinh y cosh 4 we see that we need to solve the pair of equations sinxcosh y cosh 4 cosxsinh y O 18 49 forx and y If y 0 the rst equation becomes sinx cosh 4 which cannot be satis ed by any x since sinx S l and cosh4 gt1 So y at 0 and the second equation requires that cosx 0 Thus x mz n0il 392 2 Since Sin 27 mt 1 the rst equation then becomes 1quot cosh y cosh 4 which cannot hold when n is odd If n is even it follows that y i4 Finally then the roots of sinz cosh4 are z 2nni4i n0ili2 The problem here is to find all roots of the equation cosz 2 We start by writing that equation as cosxcosh y i sin xsinh y 2 Thus we need to solve the pair of equations cosxcoshy 2 sinxsinhy 0 forx and y We note that y 0 since cosx 2 if y 0 and that is impossible So the second in the pair of equations to be solved tells us that sinx 0 or that x m n O i 112 The rst equation then tells us that 1 cosh y 2 and since cosh y is always positive It must be even That is x 2m n Oili2 But this means that cosh y 2 or y cosh 12 Consequently the roots of the given equation are z2n7ricoshquot2 n0ili2 To express cosh3912 which has two values in a different way we begin with y cosh3912 or cosh y 2 This tells us that equot equot 4 and rewriting this as e 2 4e l0 we may apply the quadratic formula to obtain e 2 iwi or y ln2 i 5 Finally with the observation that W i ln2 V51n 2N ln2 man5 we arrive at this alternative form of the roots z2nniiln2V n0ili2 50 SECTION 34 1 To nd the derivatives of sinhz and coshz we write Z 2 z z isinh de e lie e e 8 coshz dz dz 2 2 dz and d i d e e z 1 d ez e h z z cos 2 dz 2 j 2 dz e e smhz 3 Identity 7 Sec 33 is sin2 2 0052 z 1 Replacing z by iz here and using the identities siniz i sinhz and cosiz coshz we nd that izsinhz z cosh2 z 1 or cosh2 z sinh2 z 1 Identity 6 Sec 33 is cosz Z2 coszlcosz2 sinzx sinzz Replacing zx by i2l and z2 by iz2 here we have cosizl z2 cosiz1cosizz sinizsiniz2 The same identities that were used just above then lead to coshz1 22 cosh z1 cosh z sinh zl sinh z 6 We wish to show that lsinhxlSlcosh zlS coshx in two different ways a Identity 12 Sec 34 is Icosh zl2 sinh2 x cos2 y Thus lcoshzl2 sinh2 x 2 0 and this tells us that sinh2 x S Icosth2 or lsinhxls Icosh zl On the other hand since lcoshzl2 cosh2 x 1 cos2 y cosh2 x 1 cos2 y cosh2 x sin2 y we know that I cosh zl2 cosh2 x lt 0 Consequently Icosh zl2 S cosh2 x or Icoshzls coshx b Exercise 11b Sec 33 tells us that lsinh ylSlcoszlS cosh y Replacing z by iz here and recalling that cosiz coshz and iz y ix we obtain the desired inequalities 7 a Observe that e e quot e equot ez e39z e z 2 2 2 zm e an39 e s1nhz 7171 smhz 51 b Also zm39 zm39 z m z ni z z z z e e e e e e e e e e coshz m cosh z 2 2 2 2 c From parts a and b we nd that sinhzm sinhz sinhz coshz m coshz coshz tanhz m 9 The zeros of the hyperbolic tangent function sinhz tanhz coshz are the same as the zeros of sinhz which are 2 nm n 0tl12 The singularities of tanhz are the zeros of coshz or zgmti n0ili2 15 a Observe that since sinhz i can be written as sinhxcosyicoshxsiny i we need to solve the pair of equations sinhxcos y O coshxsin y 1 If x O the second of these equations becomes sin y 1 and so y g2n7r n 0ili2 Hence 1 z2n2 m n0ili2 If x 0 the first equation requires that cosy0 or y n7r n 0iliZ The second then becomes 1 39 coshx 1 But there is no nonzero value of x satisfying this equation and we have no additional roots of sinhz i b Rewriting coshz i as coshx cos y i sinhxsin y 21 we see that x and y must satisfy the pair of equations coshxcosy 21 sinhxsiny 0 52 If x0 the second equation is satisfied and the first equation becomes cosy Thus y cos i37 21271 n 0iliZ and this means that z2ni7ti n0i1i2 If x 75 0 the second equation tells us that y rm n 0ili2 The first then 1 becomes 1 coshx But thlS equation in x has no solution smce coshx 21 for all x Thus no additional roots of coshz are obtained 16 Let us rewrite coshz 2 as coshxcosy i sinhxsin y 2 The problem is evidently to solve the pair of equations coshxcos y 2 sinhxsin y 0 If x 0 the second equation is satisfied and the first reduces to cos y 2 Since there is no y satisfying this equation no roots of coshz 2 arise If x at 0 we find from the second equation that sin y O or y nil n 0tli2 Since cosnn39 1quot it follows from the first equation that 1quot coshx 2 But this equation can hold only when n is odd in which case x cosh 12 Consequently zcosh 22n1ni n0ili392 Recalling from thesolution of Exercise 18 Sec 33 that cosh3912 iln2 5 we note that these roots can also be written as z11n2 2n1m n 012 53 Chapter4 SECTION37 2 2 2 2 1 1 dt 1 1 1 o 2 a t 1 dt tz t 11 t 2 211n2 2 11n4 zl i2 quot5 m e 1 it 7 V3 z l 39 b e dt 211 2icos3 zsm3 4 4 c Since new e39bquot we find that e39quot quotb 1 1 j e39z39dt lim Je dt hm lim1 e39quot when Re z gt o o b bo z z b bu t0 3 The problem here is to verify that 2quot 0 when I J etm e meda m n 0 271 when m n To do this we write 2 e a I Jemaemadg JelMn9d0 0 O and observe that when m 9 n cumaw Tquot l l O im n o im n im n 39 When m n I becomes I 7219 27 and the veri cation is complete 0 4 First of all 13 7 Ie W dx Iequot cosxdxije sinxdx 0 0 0 But also lixdx Je li 2 quot em quot quotequotequotquot 1 equotl39l1 1e r 1equot 0 11quot 11 l i 2 0 54 Equating the real parts and then the imaginary parts of these two expressions we nd that 1equot 2 e r 1 It 1 Je cosxdx and Je smxdx 0 0 Consider the function wt equot and observe that 21 21 I ea 21 1 1 jwtdt jequotdt 7 0 o 0 1 l l 0 Since lwc27r 0 lekl2n39 271 for every real number c it is clear that there is no number c in the interval 0 lt tlt 27 such that 27 j wtdt wc27t 0 a Suppose that wt is even It is straightforward to show that ut and vt must be even Thus a jwtdt j utdt ijvdt 2 utdt 2i vtdt u a O 0 2jutdt if vtdtl 2 wtdt 0 0 0 b Suppose on the other hand that w t is odd It follows that ut and vt are odd and so I wtdt jumdt i j vtdt o i0 0 Consider the functions Pnx 371 Ni x2 c039 d6 n 012 0 where 1 S x S 1 Since xi391x2 cost9x2 1x2cos2 e stz 1x2 1 it follows that xil xzcose 1quot n 1quot lt quot pnx it was Wide 1 55 SECTION 38 1 a Start by writing I Jw tdt Iu tdt i Jv tdt b b b The substitution 139 t in each of these two integrals on the right then yields I I ut39d139 if vrdr jurdr if vrdr fw1dt39 b b a a a That is a b Iw tdt I w1d t b a b Start with I jwtdt jutdt if vtdt V a and then make the substitution t p1 in each of the integrals on the right The result is 3 19 H I jut rgt1 39ltrgtdr ii vt ltrgt1 39rgtdr I wt c1 39ltrdr That is b H J wow 1 w r1 ltrgtdr 3 The slope of the line through the points a a and b in the rt plane is b a 3 0 So the equation of that line is b a B a ta 1 0 56 Solving this equation for t one can rewrite it as b a 1 aB ba B a 3 0 1 Since t 7 then b a 17 0 aBba 257 6 0 If Z Z T where 20 W i t and t 17thcn 27 x 139 iy 139 Hence Z 139 x 7 iy 7 x39 1 39 i 17 x39 139 iquotl f 7 z39 T 39T If WU f 20 and f Z 3600 iVx y zt xt iyt we have Wt uxtYCt ivxtyt The chain rule tells us that du d I I BEL14 uyy and 7vxx vyy and so w t uxx uyy39 ivx vyy39 In view of the Cauchy Riemann equations u vy and u v then w t uxx y i vxx39 uxy u iVx iy That is W t uxtytl ivxtytlx t iy39t f 39ztlz39t when t to 57 SECTION 40 l a Let C be the semicircle z 2e 9 0 S 6 S 7 shown below 2 0 2 x Then z2 2quot 2rl jc z dz L1dz l2 ei21e9d6 2 g e 9 1d6 i9 quot 2if e 2ii7ri 427ri l 0 b Now let C be the semicircle z 2e 7 s 0 S 27 just below This is the same as part a except for the limits of integration Thus 9 2r Icz2dz2ie9 2i i27ri 7r42 39i z 1 If C Finally let C denote the entire circle z 2e 0 S 9 S 27 In this case I E dz 4717i C z the value here being the sum of the values of the integrals in parts a and b 2 a The are is Cz1eie MS 65271 Then 2quot i 27 I em 21 Ice 1dz J 1 a 1iequot d9 i Veda 37 1 in it 1 5e4 e2 21 10 58 b Here Czx0x2 Then JCZ1dZJxldx 22x0 3 In this problem the path C is the sum of the paths C1 C2 C3 and C4 that are shown below The function to be integrated around the closed path C is fz we We observe that C C1 C2 C3 C4 and find the values of the integrals along the individual legs of the square C i Since C1 is zx0SxSl 1 12 xx It Ll Ire dz nge dx e 1 ii Since C2 is z 1iy0y1 l 39 l 39 JG Irequotde III e quot ia39y Kiri e quot dy 2e 0 0 iii Since C3 is z 1 xi 05x 5 1 I n l j e de 7r e 1 gtquot 1dx M e39 dx equot 1 C3 0 0 iv Since C4is z ily 0 S y 51 l 1 Ic e zdz leNU l iJein ydy 2 0 Finally then since I Irequotde I e dz J39 ne dz J xe dz J nezzdz c C1 c2 6 c we find that L ne dz 4equot 1 59 4 The path C is the sum of the paths C1zxix31SxS0 and C2zxix3OSxS1 Using fZ10nC1 and fz4y4x30nC2 we have 0 l C fzdz jcl fzdz C1 fzdz 11 i3x2 dx 4x31i3x2dx idx 3iix2dx 4jx3dx 12ijx5dx 1 1 0 0 xE1 ix3f1 x4 24 1i12i23i 5 The contour C has some parametric representation 2 zt a S t S b where za z1 and zb zg Then b Ldz j z39tdt zt zb M 22 21 6 To integrate the branch z39m e39l l lzlgt 0 0 lt argz lt 27 around the circle C z e O S 0 S 27 write 21 21 21 IC z ll dz J39Ce ltlogz dz J39e ltlnl10ie19d9 iJ e199 grade iJeGdg eZ 0 0 0 7 Let C be the positively oriented circle Izl l with parametric representation z e O S 0 S 271 and let m and n be integers Then 21 21 L zmzndz eia mei9 fad9 iJ emeweinsda 0 But we know from Exercise 3 Sec 37 that 2 m M 0 when mam Je e d0 o 271 when m n 60 Consequently m 0 when m1 n Iz z dz C 2m when m1n 8 Note that C is the righthand half of the circle 22 y2 4 So on C x x4 yz This suggests the parametric representation C 2 14 y2 iy 2 s y 5 2 to be used here With that representation we have deziW 2 2 J yydyiJ 14 y2dy 2 in24 yzdy4iJ2 dy 4isin39ll2 2 34quot 2Q4 2 2 2 y 4isinquot 1 sin 1 1 4139 g 325 4m 10 Let Co be the circle 2 z0 Re 71 S 6 S 72 dz 1 9 a CozwzoiRelee dG z Ld92m b When nili2 Lo 2 20 quot dz r12equotquotquot l RieiadB iR feiwde x u eimr eimr l 2Rquot sin mt 0 11 In this case where a is any real number other than zero the same steps as in Exercise 10b with a instead of n yield the result 2R sina7t a Lo z zo quotdz i 61 12 a The function fz is continuous on a smooth arc C which has a parametric b representation 2 21 a S t S 1 Exercise 1b Sec 38 enables us to write 1 Ifztz tdt 1fzrgt1z39 r1 39rdr where a a 21 z But expression 14 Sec 38 tells us that z r r Z r and so b I3 I fztz tdt 1 rm 139Z 1d1 Suppose that C is any contour and that f z is piecewise continuous on C Since C can be broken up into a finite chain of smooth arcs on which fz is continuous the identity obtained in part a remains valid SECTION 41 1 Let C be the arc of the circle z 2 shown below Without evaluating the integral let us find an upper bound for I To do this we c z 1 note that if z is a point on C z2 1 2 lull1 Iz2 1 i4 1 3 Thus zzl lz2 1 3 62 1 Also the length of C 18 247r 7 So taking M and L 71 we find that d 5ML1 3 2 The path C is as shown in the figure below The midpoint of C is clearly the closest point on C to the origin The distance of that midpoint from the origin is clearly g the length of C being 5 i Z4 1 s 4 lzl 2 Hence if z 18 any pomt on C lzl Z This means that for such a pomt Consequently by taking M 4 and L 5 we have sML4V dz leg 3 The contour C is the closed triangular path shown below Le Zdz we let 2 be a point on C and observe that le 2 Sle l2 e 1 x2y2 To nd an upper bound for 63 But e 51 since x lt O and the distance yxz y2 of the point z from the origin is always less than or equal to 4 Thus Iez Zl lt 5 when 2 is on C The length of C is evidently 12 Hence by writing M 5 and L 12 we have Ce zdz 5 ML 60 Note that if z R R gt 2 then Izz2 1 2lz2l2R2 1 and Iz 5z2 41 Iz2 1 Iz239 4 2 IzI21 llzI2 4l 122 1R2 4 Thus 2z21 l222ll lt 2R2l z 5z24 Iz 5224R2 1R2 4 when lz R R gt 2 Since the length of CR is 7tR then 2i J 222 1 lt 7rR2R21 R R2 an45224 quotR21R24 1LX11 R2 R2 and it is clear that the value of the integral tends to zero as R tends to in nity Here CR is the positively oriented circle lz R R gt 1 If z is a point on CR then Logz 2 Z llnRilltlnRlt1tlnR R2 R2 R2 since 71 lt O S 7 The length of CR is of course 273R Consequently by taking nlnR R2 M and L27rR 64 weseethat J Lo gzdzls 2 nlnR C R Since nlnR lR 11m 0 RMO R R bu 1 itfollowsthat Logz lei IEICR 22 d 0 Let Cp be the positively oriented circle lzl p O lt p lt1 shown in the figure below and suppose that f z is analytic in the disk lzl S 1 We let 2 represent any particular branch 1 2 exp I logz exp lln r 19 1 exp i r gt O a lt 0 lt 05 27 2 2 F 2 of the power function here and we note that since fz is continuous on the closed bounded disk lzl S 1 there is a nonnegative constant M such that l f zllt M for each point z Cy zquot zfzdz in that disk We are asked to nd an upper bound for To do this we observe that if z is a point on Cp lz 12fzzI2 I 3 Since the length of the path CI is 27rp we may conclude that M z 1nfzdz lt 2n39 2nM LP p p xP Note that inasmuch as M is independent of p it follows that lim p 0 C z 1 2fzdz 0 65 SECTION 43 1 The function zquot n 012 has the antiderivative 2quot n 1 everywhere in the nite plane Consequently for any contour C from a point 21 to a point 22 1 z L zquotdz Izquotdz Zr 1 22 n1 1 n z2 zln 1 23H znl n1 n1 n1 n1 I 1 52 4 b n2i z d 2 z 1t2i 2 n I 2ei2 e J39cos2 2 sm 20 sm2 1 2i 0 iei1r2el emZe 3 3 1 1 3 c 2 2dz 4 4 0 Note the function z 20 1 n ili2 always has an antiderivative in any domain that does not contain the point z 20 So by the theorem in Sec 42 Lo 2 zo quot 1 dz 0 for any closed contour CD that does not pass through 20 Let C denote any contour from z l to 2 1 that except for its end points lies above the real axis This exercise asks us to evaluate the integral l I Izidz l where 2quot denotes the principal branch z expiLogz Izl gt o 7 lt Argz lt 7 66 An antiderivative of this branch cannot be used since the branch is not even de ned at z 1 But the integrand can be replaced by the branch z expilogz Izl gt O 12rlt argz lt since it agrees with the integrand along C Using an antiderivative of this new branch we can now write zil 1 1 1 l 1 m 1 I I I I 1 1 39 5 08 103 ind i1k i1k e 1 1 ei1ln1r 0 ei1lnli7r z in 1 e z 11 i1 i1 1i l i 1 e li 2 SECTION 46 2 The contours C1 and C2 are as shown in the gure below In each of the cases below the singularities of the integrand lie outside C1 or inside C2 and so the integrand is analytic on the contours and between them Consequently Jgfk hJgfu h b When f z 67 l a When f z m the smgulanties are the pomts z i i z2 the sin ularitiesare t 2 0391i2 sinz2 g a z quotMquot c When fz the singularities are at z 2m n 01Jr2 a In order to derive the integration formula in question we integrate the function e 2 around the closed rectangular path shown below y abi A abi V V A L a 0 f 4 x Since the lower horizontal leg is represented by z x as x5 1 the integral of equot2 along that leg is its 4x 2 equot dx a 0 Since the opposite direction of the upper horizontal leg has parametric representation 2 x bi a S x S a the integral of equot along the upper leg is u a I a 391 1 2 z z 1 2 Ie quotquot0 dx equot Is quot e dx eb Je costxdxHequot Ie quot stbxdx 1 u quota 3 orsimply 2equot2 I 2quot2 cos 2bxdx 0 Since the right hand vertical leg is represented by z a iy O S y S b the integral of e39z along it is b b 2 1 2 J e n0 ldyle a J39ey e lZaydy 0 68 Finally since the opposite direction of the lefthand vertical leg has the representation z a iy o s y s b the integral of e quot along that vertical leg is b b z I I z 2 Je 9 zdy te quot Ie e39mdy 0 0 According to the CauchyGoursat theorem then a a 39 39 H b b 2 equot2dx Zebz J equot cos bedx if Ie ze39izqdy ie39 z J e ze mdy 0 0 A o i o I p o I and this reduces to i v 4 V a a b Iaquot cos 2bxdx e quot2 Ie zdx o2quotquotz 1 jequot2 sin 2ay dy 0 0 0 b We now let a gt oo in the final equation in part a keeping in mind the known integration formula and thefactthat z 1 b 2 2 2 b 2 e 0 5 Ie sm2aydy Se H Ie dy gt0 as a gtoo 0 0 The result is J bgt0 Jequot1 cos 2bxdx le39bz 0 2 6 We let C denote the entire boundary of the semicircular region appearing below It is made up of the leg C from the origin to the point 2 1 the semicircular are C2 that is shown and the leg C3 from z l to the origin Thus C C1 C2 C3 69 We also let f 2 be a continuous function that is de ned on this closed semicircular region by writing f O 0 and using the branch fzei92 rgtO 12r lt 9lt 32 of the multiplevalued function 2 The problem here is to evaluate the integral of f 2 around C by evaluating the integrals along the individual paths 6 C2and C3 and then adding the results In each case we write a parametric representation for the path or a related one and then use it to evaluate the integral along the particular path i C1 zrequot OSrSI Then If 2 1 2 d w m 112er r 3r 0 3 ii C2z1e 0lt0S7t Then jc fzdz j 43quot 2 ie d6 i j e 39 2d6 ie 3m 3H 1 0 i z 0 o l o 3 3 iii C 2 re 0 S r 51 Then 2 z 3 0 l 1 1 CS fzdz Cl fzdz Je quot 2 1dr i j Fair iB rm The desired result is Icfzdz J fzdz IC1 mm C3 fzdz 1 i 34 o The CauchyGoursat theorem does not apply since f z is not analytic at the origin or even de ned on the negative imaginary axis SECTION 48 1 In this problem we let C denote the square contour shown in the gure below 2139 A 70 e39zdz a 2m e z mm 2m39 i 27 cosz coszz2 8 cosz ni 2 39 0 JCzz28dz L 2 0 dz 7 228 20 27 8 4 zdz z2 1 m39 d 2 2 C J39C221 jCz12 Z m 2 z l2 m 4 2 3 coshz coshz 2m EI m d L Z4 dz JC z03 dz 3 dzacoshzzo O O tanz2 tanz2 zggg i z e L zxo2 dz Jczxol1 dz tan2zxo 27rilsec2 53 inseczx when 2 lt x0 lt 2 2 2 2 2 Let C denote the positively oriented circle Iz il 2 shown below a The Cauchy integral formula enables us to write dz dz lz2i 1 J 71 d 2 2m JG z2 4 I0 z 2iz2i L3 z 2i z m z2i F2 4i 2 b Applying the extended form of the Cauchy integral formula we have 1 dz 4 dz 1 l cz242 Cz Zi2z2i2 C zZiY 1 dZZ2i2 z2i 2 47ri 47ri It WWz2i 40 164i 16 71 3 Let C be the positively oriented circle lzl 3 and consider the function ZZZz Z zw 300 L dz lwlat 3 We wish to find gw when w 2 and when M gt 3 see the gure below 0 E We observe that 2 2 2 82 JC Z z2 dz 2m2z2 z 2z2 2M4 3m On the other hand when lwl gt 3 the CauchyGoursat theorem tells us that gw 0 5 Suppose that a function f is analytic inside and on a simple closed contour C and that 20 is not on C If 20 is inside C then 1quot7dz2 fzdz fzdz 2 c zzo mf 20 and Iczzo2 Czzoll 139 0 I f 39zdz f zdz Cz CQaf39 The CauchyGoursat theorem tells us that this last equation is also valid when a is exterior to C each srde of the equation being 0 7 Let C be the unit circle z e ns as it and let a denote any real constant The Cauchy integral formula reveals that Jazz dz JG ze0 dz 2meazzeo 271i 72 On the other hand the stated parametric representation for C gives us e quotexpaei9 is z sz r Tle d9 1Jexpacos 6 1 sm 6d9 i I emwemmede iJequot cosasin 6 i sinasin 0d6 J em sinasin 6d6 i I em cosasin 0d9 7t e Equating these two different expressions for the integral JC dz we have z j em sina sin 9d6 i em cosasin 9d9 2m Then by equating the imaginary parts on each side of this last equation we see that Jews cosasin 0d6 27 and since the integrand here is even Jew cosasin 9d6 7 o 8 a The binomial formula enables us to write 1i 2 quot 421quot n 2n2k k Pquotzn2quotdzquotz 1 n2quotdz 2kz D39 k0 We note that the highest power of z appearing under the derivative is 22quot and differentiating it n times brings it down to zquot So P39 z is a polynomial of degree n b We let C denote any positively oriented simple closed contour surrounding a fized point z The Cauchy integral formula for derivatives tells us that dquot 2 n n s2 1quot 1 Ol2 dzquot z 271 L s zquot1 ds n Hence the polynomials P39 z in part a can be written 1 32 1quot Pz ijc Pam ds n 012 73 c Note that s2 1quot s lquots1quot slquot s1nl slnl 3 1 39 Referring to the nal result in part b then we have 1 sz lquot 1 1 slquot 1 P 1 n n 2 Cs11ds 22 Ic s1 ds 2quot2 1 n 012 Also since s2 1quot s 1quots 1quot s 1quot s1quot1 s 1quot1 s1 7 we have 1 s2 1quot 11 31quot 1 n n Pn 1 2nl i CS1nldg 2n s1 ul quot02112vquot We are asked to show that f 3619 sz3 u i f 0 7 a In view of the expression for f 39 z in the lemma f39zAzf39z1I 1 1 How Az 2m c szAz2 s z2J Az 1 2 J s z AZ 2fSdS 2m CszAz2s z Then f39zAzf39zi fsds l 2s zAz 2 Az mittz3 27 quot6zAz2sz2 sZ3fsw 1 33 zAz 2Az2 27ti l s z Az2s z3 fmds39 74 b We must show that 3DIAzl 2mzl2 M I 3s zAz 2Az2 C d lAzl2d3 szAz2sZ3 f sds 5 Now D d M and L are as in the statement of the exercise in the text The triangle inequality tells us that 3s zAz 2Az2Is 3s zl lAzl 2IAzlzs 301M 2Az2 Also we know from the veri cation of the expression for f 39 z in the lemma that Is z Azl 2 d Azgt O and this means that ls z Az2s z3l 2 d Azl2d3 gt 0 This gives the desired inequality Ifwe let Az tend to 0 in the inequality obtained in part b we nd that 3s zAz 2Az2 1 A23 lcm ds 0 This together with the result in part a yields the desided expression for f quot z 75 Chapter 5 SECTION 52 1 We are asked to show in two ways that the sequence 1quot zquot 2ln2 n 12 converges to 2 One way is to note that the two sequences xquot 2 and yquot quot1 n 12 n2 of real numbers converge to 2 and 0 respectively and then to apply the theorem in Sec 51 Another way is to observe that lzl 2 Thus for each 8 gt 0 n lzl 2 lt 8 whenever n gt no 1 where no is any posmve mteger such that no 2 s 2 Observe that if z 2 111 n 12 then quot2 1 r lzl 4 4 gt 2 n But since 92quot Argzon 7t and 92 Argon gt 75 n 12 the sequence 9 n 12 does not converge 3 Suppose that lim 2 z That is for each s gt 0 there is a positive integer no such that N lz zllt 8 whenever n gt no In view of the inequality see Sec 4 lz zl 2 Ilzol lzll it follows that llznl lzlk 8 whenever n gt no That is limlznlzl nn The summation formula found in the example in Sec 52 can be written when zlt 1 If we put 2 rem where 0 lt r lt1 the lefthand side becomes 2rei9quot Zrquoteiquot9 2r cosnG i2 rquot sin n6 711 111 nl n1 and the righthand side takes the form re 9 1 re39 9 rem r2 rcos 9 r2 irsin 9 l rew l re39i 9 l rei9 equotquot r2 l 2r0089r2 Thus we on 2 I rcosGr rsrnQ 2rquotcosn0l2r smn9 2 2 quot1 quot1 1 2rcosOr 1 2rcosGr Equating the real parts on each side here and then the imaginary parts we arrive at the summation formulas rsin 9 2 on rcosG r and Zrquot s1nn6 2 l 2rcos9r 2r cosn6 2 1 2rcos9r quot1 n1 where 0 lt r lt1 These formulas clearly hold when r 0 too Suppose that Zzn S To show that 22 3 we write zl xI iyn S X iY and 131 n1 appeal to the theorem in Sec 52 First of all we note that ix X and fly Y nl nl Then since 29yquot Y it follows that nl if in iy ix i y XiY 3 n1 nl n1 77 8 Suppose that 22 S and 2yl T In order to use the theorem in Sec 52 we write nl nl znxuiyn SXiY and wuuniv TUiV Now 2x X 2y Y and Euquot U 2v V nl nl nl nl Since 20quot uXU and 20 vn Y V 111 111 it follows that 2mquot uiy v X UiY V nl That is 2m iyn u iv XiY UiV n1 or 2g Wquot ST nl SECTION 54 1 Replace 2 by z2 in the known series 0 211 z cosh l lt oo 2 2n Z to get a 4n h 2 z I lt cos z g0 Z Then multiplying through this last equation by 2 we have the desired result an z4nl zcoshzz z zlt co n0 78 2 b Replacing z by z 1 in the known expansion ez 25 lzllt co we have n equot1 zlt oo 80 e ez39le egg alg lzllt 00 2 Izlltl as well as its condition of validity to get 1 1quot 4n 1z49 32quot Z Md Then if we multiply through this last equation by 3 we have the desired expansion fz i 1 lzl lt 4339 32n2 n0 6 Replacing 2 by z2 in the representation 2nl quot z 00 San 1 am lzllt wehave 2 as quot z4n2 smz 01 am Zlt 79 Since the coef cient of zquot in the Maclaurin series for a function f z is f quot0n this shows that forum 0 and flt2quot00 o n 01 2 The function 1 12 has a singularity at z 1 So the Taylor series about 2 i is valid when lz il lt 45 as indicated in the gure below To nd the series we start by writing 1L1L 1 z 1i z i l i l z i1 i39 This suggests that we replace 2 by z i 1 iin the known expansion 1jz lzllt1 1i 1 3quot Iz lt The identity sinhz m sinhz and the periodicity of sinh z with period 2m tell us that sinhz sinhz xi sinhz ni So if we replace z by z m39 in the known representation an 2nl smhz IZIlt 80 and then multiply through by l we nd that 2n1 zm sinhz quot0 2n 1 lz m lt no 13 Suppose that O lt Izl lt 4 Then 0 lt Iz 4 lt1 and we can use the known expansion Mg 1 1z z lzlt1 u c To be speci c when 0 lt Izl lt 4 1l l1 zquot z 39ll quot2quotquot 1 quotzquot 42 2 39T 4 22Z n0 n0 ul quot0 SECTION 56 1 We may use the expansion 2211 sinz l n I 00 E 2n1 ZK to see that when 0lt Izl ltoo zzsm1 1quot 11i 1quot ml z2 quot0 2n1 z n12n1 z 3 Suppose that lltlzllt co and recall the Maclaurin series representation izquot I lt1 1Z quot0 z This enables us to write 1 1 1 1 1 quot 1 1 l Z In z 2 lt gt Replacing n by n 1 in this last series and then noting that 1quot391 1quot l2 1 81 we arrive at the desired expansion 1 z 5 Zquot z 4 The singularities of the function fz 2 11 are at the points z O and z 1 Hence 2 2 there are Laurent series in powers of z for the domains 0 ltlzlt1 and 1ltlzllt co see the gure below To nd the series when 0 lt zlt I recall that 11 22quot Izlt l and write 2 n0 1 1 1 quot 1 1 39 1 1 fz z z 2 z zquot 221z 22 2 z2 z E 2 z z Asfor the domain 1ltzllt co note that llzl lt1 and write fz 1 1 iquot1quotizin zsn0 2 n0 n3 5 a The Maclaurin series for the function z is valid when izllt 1 To nd it we recall z the Maclaurin series representation 1zz Izllt1 1 2 quot0 for 1 and write l z Z1 1 a n n1 a n zl z1ZZ Xz 22 21 l z n0 n0 n0 22 jz 422 IzIlt 1 nl n0 n1 82 b To find the Laurent series for the same function when 1ltlzlltoo we recall the 1 Maclaunn senes for 1 that was used 1n part a Smce lt 1 here we may wnte z 1 1 z1 z I l 1 1quot quot 1 39 1 z1 1l 2 11 2 z 252quot z 2 quot 1 39 1 1 12 lltllltltgto g2quot E2quot 242 z 1 7 The function fzm has Isolated Singularities at 20 and zii as indicated in the gure below Hence there is a Laurent series representation for the domain 0 ltzlltl and also one for the domain 1ltzlt oo which is exterior to the circle z1 39 A a 14 4 To nd each of these Laurent series we recall the Maclaurin series representation lzlt l 1 n 17 22 For the domain 0 ltzlt l we have fz 1 2 lizzquot 1n Z2111 1 i1nz2n l i1nlz2nl 1 z 1z z 110 110 Z quot1 quot0 Z On the other hand when 1ltzlt co 1 1 1 1 quot 1quot 1quot1 Z 39 f ZS 1 Ji Z3 quot0 22 g zzn3 quot214 z2nl In this second expansion we have used the fact that lquot391 1quotquot 12 1 8 a b 10 a 83 Let a denote a real number where l lt a lt 1 Recalling that 1 N L z 22quot Izllt 1 n0 enables us to write a 1 1 j 1nl z a z 1 az oz or a aquot z a 2 lalltlzllt 00 nl Putting z e on each side of the nal result in part a we have a iaquote nl h9 e39 a But a a cosO aisin6acosB a2 iasin0 em a cosO aisin0 cosG a isin6 1 2acos a2 and 2 aquotequotquotquot 2aquot cosnB i2 a sinnB nl nl n1 Consequently asine 1 2ac056 a2 acosG a2 and 2 quotSmquot9 nl 2aquot cosn6 nl when 1ltalt1 Let z be any xed complex number and C the unit circle w equot 11 S q S 7 in the w plane The function f w expZw 2 w has the one singularity w O in the w plane That singularity is of course interior to C as shown in the gure below 84 I w plane Now the function f w has a Laurent series representation in the domain 0 ltlwllt oo According to expression 5 Sec 55 then epr w i Jl zwquot 0ltWllt 2 w nen where the coefficients 1 z are 1 expw 1 L Jz n M C M dw n 012 Using the parametric representation w e 7 5 I lt 7 for C let us rewrite this expression for J z as follows 1 1r exp e e39 1 1r i 1 12 2m I em 1e d 2m rexphzsm e dd That is Jz 2 Irexp in zsin d n Oili2 b The last expression for Jl zin part a can be written as Jnz 517 cosn zsin isinn zsin d Z lncosmp zsin d 2 jsinn zsin d 1 quot i 2 26cosn zsm wrap0 n o12 85 That is Jz jcosmcp zsin d n Oi1t2 0 11 a The function f z is analytic in some annular domain centered at the origin and the unit circle C z e 7 S 4 lt n is contained in that domain as shown below N For each point z in the annular domain there is a Laurent series representation fz ianz 2 n0 nl where 1 fzdz 1 quotfe i 1 x i in anE JCT imle nigh Ei e39k 39d quot 091 2quotquot and 1 fZdZ1 fe i 1 u w bn RTICW 7E ei nlle d p z ir i e e d quot 1 2 quot39 Substituting these values of a39 and bI into the series we then have n0 1quot quot fz 25mm quot d zquot fequote39quot39d fz ginIMW ffltequot ef quot JIM Ill 1 86 b Put 2 em in the nal result in part a to get ew i eiqw gi eu eman ein9 d i 1 n i 1 u i fe 9 5 Me 39d 72 Me cosn0 gt1d z 1 1 If u6 Re f em then equating the real parts on each side of this last equation yields 1 1 It um Eu d n2jxu gtcosrnlte 1d SECTION 60 1 Differentiating each side of the representation 1 E 22quot IZK 1 we nd that quot0 1 1 a n a i n a l n 41 H I i lz2 dzgz gdzz Em go 2 Z Another differentiation gives 2 1 n n a 1 n n quot quot l z dzEquotlz 2quot1z n2quotquot1z 1 nln2z lzlltl n0 u0 2 Replace z by 1 1 2 on each side of the Maclaurin series representation Exercise 1 1 3 im 1zquot Izllt 1 V as well as in its condition of validity This yields the Laurent series representation 1 quot1quotnl 1 cc zz 1 I lltlz lt 87 3 Since the function fz1 2 has a singular point at z 0 its Taylor series about 20 2 is valid in the Open disk Iz 2lt 2 as indicated in the gure below To nd that series write 1 l 1 1 z 2z2 2 lz22 to see that it can be obtained by replacing z by z 2 2 in the known expansion 1 quot E gz lzllt1 Specifically 11 Z 2 quot 1 2 lz 2lt2 0139 1 quot 1quot g 2 H z 2 lz 2lt2 Differentiating this series term by term we have 1 In quota u 1 n1 2 quot211222 12 12 n1z 2quot lz 2Ilt2 Z quot1 2 n0 2 Thus 1 1 39 2 2 quot 1quot 1 2 Z 42 n 2 lz 2lt 4 Consider the function defined by the equations ezl h 0 2 z wenz 1 when 2 0 88 When 2 0 f z has the power series representation 1 z z2 23 z 22 fZ l39l 2 39 quot39 l 1 E Nu Since this representation clearly holds when z 0 too it is actually valid for all z Hence f is entire Let C be a contour lying in the open disk lw lllt1 in the w plane that extends from the point w 1 to a point w 2 as shown in the gure below 0 w w plane According to Theorem 1 in Sec 59 we can integrate the Taylor series representation 1 Z lquotw 1quot lw 1lt 1 W n0 term by term along the contour C Thus dw e n n w n 71 C7 2534 w 1 dw EH Cw 1 dw But Id w Logwf Logz Logl Logz C w 1 w and Z n1 z n1 Jw 1quot jw 1quotdw W 1 z 1 C 1 n 1 1 n 1 Hence 1quot n1 1quot391 n L 1 1 I 1llt1 ogz gn o n z gt z gt and since 1quot391 1quot 12 1 this result becomes so n1 Logz 2 1 z 1quot lz1lt1 n n1 89 SECTION 61 1 The singularities of the function fz f zz 1 nd the Laurent series for f that is valid in the punctured disk 0 ltzllt 1 shown below Z are at z 0 i i The problem here is to We begin by recalling the Maclaurin series representations 2 3 z z 2 2 00 e l123 lzlt and 1 I zlzzzz3m lzlt1 which enable us to write 2 1 2 1 3 e 1z z I lltoo 2 6z 2 and 1z2ZA zsu l 22l Multiplying these last two series term by term we have the Maclaurin series representation Z 1 l 1 2 3 1 Z zz 6z m 1 5 1 2 3 z 2 z 6 2 which is valid when zllt l The desired Laurent series is then obtained by multiplying each side of the above representation by 1 z e 1 1 5 39 1 2 0lti lltl zzzl z 2 z We know the Laurent series representation 1 1 7 0ltI Ilt7r 360Z z from Example 2 Sec 61 Expression 3 Sec 55 for the coefficients b in a Laurent series tells us that the coef cient b1 of l in this series can be written z 1 I dz 1 2m czzsinhz where C is the circle z 1 taken counterclockwise Since bl i then dz I m39 271 It z2 sinhz l 6 3 The problem here is to use mathematical induction to verify the differentiation formula n fzgz quot 2031 zg quotquot z n 12 k0 The formula is clearly true when n 1 since in that case it becomes f 28z39 f zg39z f 39zgz We now assume that the formula is true when n m and show how as a consequence it is true when n m 1 We start by writing f 23z quot fzgz quot f zg z f zgzl quotquot f zg39Zl39quot f Zgzlquotquot if gtltzgltquot quot ltz i f ltzgtg ltzgt k0 150 Ms mjf zg quot39 quotzquot m fquotquotzgquotquotquot quotz k k k l 7quot II 0 f zgquotquot z kn 1 f 0quot zg quotquot z f quotquot zg2 kl 91 But a 7 mm m m m1 m1 k k l km k k lm k1 km1 k k and so m1 fzg2quotquot fzg quot39 z2 k f zg quot 39quot zf quot zgz m kl m1 l fzgz quot39 ifquot f zg quot 39quot z The desired veri cation is now complete We are given that f z is an entire function represented by a series of theform fzzagz2 agza Izllt co a Write gz f f z and observe that firmgltoz922 25m lzltoo It is straightforward to show that 839Z f 39f 2f 39Z g z f fzf zz f39fzf z and g z f quot39f Zf 39Z3 2f Zf Zf fZ f quotf Zf Zf quotZ f f 1f Z Thus 30 0 g 0 1 g 0 4a and g 0 12az2 as and so ffz z 207 2a22 a3z3 lzllt co 92 b Proceeding formally we have ffz fz a2fZ2 asfz za2z2 a3z3 a2z 12z2 a323t2 cz3za2272 03233 za2z2 c13z3azz2 2a z a3z3 z Zazzz 2a22 a3z3 c Since sinzz 3 z0z2 lz3 lzllt co 3 6 the result in part a with a2 0 and a3 tells us that sinsinz z1zg zlt 8 We need to find the rst four nonzero coef cients in the Maclaurin series representation 1 W2 coshz on 2 39 This representation is valid in the stated disk since the zeros of coshz are the numbers 2 gImr n Oili2 the ones nearest to the origin being z tati The series contains only even powers of 2 since coshz is an even function that is Eml 0 n 01 2 To nd the series we divide the series 2 4 6 z z z 12 14 1 6 cosh 1 1 llt z 2 4 6 zz 242 7202 Z into 1 The resultis 1 1 5 61 7 1 zz 4 6 Izllt coshz 2 242 7202 1 1 2 5 4 616 coshz 212 412 6Z Since 1 E2 E E E2 44 66 coshz 2Z4z6z thistcllsusthat 1301 E2 1 E45 and E66l 93 02kg M 94 Chapter6 SECTION64 1 a Letus write 1 1 1 1 2 3 1 1 zz z 1 2 0ltI llt1 zzzz1z z z zz z The residue at z O which is the coefficient of 1 is clearly 1 z b We may use the expansion 2 4 6 Z 5Z cc cosz1 24 61 lzlt towrite 239 z2 4 z4 639 26 239 z 439 23 639 zs 0ltlzlltoo The residue at z 0 or coefficient of l is now seen to be 2 z 6 Observe that z sinz 1 1 z3 zs 22 24 z smz z z 0ltzlltoo z z z35 35 Since the coeffiCIent of in this Laurent senes is 0 the resxdue at z 0 IS 0 z 1 Write cotzi cosz Z4 24 sinz andrecallthat 7 4 2 4 cosz1 z1zz Izlltoo 2 4 2 24 and 3 5 3 5 sinzzz z zz lzlltoo 95 Dividing the series for sinz into the one for cosz we nd that Oltzllt 7r 5 Oltlzlt7t Note that the condition of validity for this series is due to the fact that sinz 0 when c t z mc n 0i1t2 It is now ev1dent that of has residue Zl 5 at z 0 z e Recall that 23 z5 smhzz3 5 lzltoo and 1 2 1zz zltoo lz There is a Laurent series for the function sinhz 1 1 smh z41z2 Z4 that is valid for O ltlzllt 1 To find it we first multiply the Maclaurin series for sinhz 1 1 2 and 2 l 1 1 inh 3 5 H 2 44 s z1zz z 6z 1202 12 z 6 120 z3 z5 ZS zza 0ltlzlt1 96 Wethenseethat sinhz l 71 Z4lz2 Z3 6 z 0ltzllt1 2 In each part C denotes the positively oriented circle lzl 3 a To evaluate LEE 53211 we need the residue of the integrand at z O From z the Laurent series 2222 5 L51 L11 1 z Z2 ZZ 139 2 339 z2 1z239 3 39 lt39Z39lt we see that the required residue is 1 Thus I ex Z de mlp zm C z 6 Likewise to evaluate the integral L 22 exp 1dz we must find the residue of the z integrand at z 0 The Laurent series zzexpl221llii Z 122z 3z 412 2211111 z1121312 42z2 which is valid for 0 ltlzlt co tells us that the needed residue is Hence 2 1 1 7L i ICZ 6XpdZ 27UE 3 97 l d As for the integral JC 2 zdz we need the two residues of z1 21 zzZz zz 2 One at z 0 and one at z 2 The residue at z 0 can be found by writing zt1z llll1 222 z z2 2 z 1z2 which is valid when 0ltlzllt 2 and observing that the coefficient of l in this last z product is mg To obtain the residue at z 2 we write z1 z23 1 1 3 1 zz 2 z 2 2z2 2 22 lz 22 2 l131dz2m 2 z 2 2 22 which is valid when 0 ltz 2llt 2 and note that the coef cient of 2 in this product z is Finally then by the residue theorem z1 1 3 2 2 39 JCzZ szz m 2 2 7quot In each part of this problem C is the positively oriented circle z 2 5 z 3 then 2 1 a 39If fz l 1 l 1 2 l 1 1 1 1za 6 z zzfz z7 z4 z4 123 z z z z when 0 ltlzllt1 This tells us that chzdz 2m39Res 2m39 l 2m z z 98 b When fz 1zzwe have 1 1 1 1 l 2 4 I szL 12 122 z z 0ltzlltl 12 21110 0 Jfzdz szes C z0 z c If f z 1 it follows that f Evidently then 2 z z z jfzdz2m Res 27 C zo z z 4 Let C denote the circle Izl 1 taken counterclockwise 039 II a The Maclaurin series ez lzllt 00 enables us to write 11 110 39 1 zquot 1 1 JCCXPZ dz Icezeuzdz L e ng dz JG 2 cxpdz b Referring to the Maclaurin series for e2 once again let us write 1 a n 1 k z ex zquot zquot n0l2 132 Mk 2quot 2 Now the l in this series occurs when n k 1 or k n1 So by the residue theorem l 1 L2 exp Zsz 2mm n 012 The nal result in part a thus reduces to 1 1 ICCXPZzdz n0 99 5 We are given two polynomials Pzaoalzazzzazquot ana O and Qz b0 bIZb2z2 39bmzm but 0 where m 2 n 2 It is straightforward to show that 1 Pl z aozquotquot2 alzquotquot3 azzm394 anzmnz 0 z2 Ql z boz quot blzmquot bzzmz hm z Observe that the numerator here is in fact a polynomial since m n 2 2 0 Also since bquot 0 the quotient of these polynomials is represented by a series of the form dodlzdzzza Thatis 1 2 d d d r 22 902 0 lz 2z 0ltlzlltR2 and we see that has residueO z0 Suppose now that all of the zeros of Qz lie inside a simple closed contour C and assume that C is positively oriented Since Pz Qz is analytic everywhere in the nite plane except at the zeros of Qz it follows from the theorem in Sec 64 and the residue just obtained that JCPzdz2m39ResL Pllz2m39OO Qz z 22 Q0 2 If C is negatively oriented this result is still true since then Pz Pz d Liz 5 z IcQz z 039 SECTION 65 1 a From the expansion 3 e 1 Izlltoo we see that Z Z 100 b c d e 1 The puncrpal part of zexp at the Isolated singular point z 0 is then z 1 l 1 1 T 2 z 3 z and z 0 is an essential singular point of that function 22 1 2 involves powers of z 1 we begin by observing that The isolated singular point of is at z 1 Since the principal part at z 1 z2 Z12221z12 2z11 This enables us to write z2 zl2 2zll 1 1 2 1z z1 z z 1 39 Since the pnncrpal part IS l the pomt z 1 IS as1mple pole z The point z 0 is the isolated singular point of 51 115 and we can write z 0 ltIzllt no The principal part here is evidently 0 and so z 0 is a removable singular point of the sinz function z The isolated singular point of w is z 0 Since 2 2 4 3 2931 1543 lzz 0ltzltoo z z 2 4 z 2 41 1 cosz the prmcrpal partis This means that z0 is a31mple pole of 2 Upon writing 1 i we find that the rinci a1 art of 1 at its isolated singular point z 2 is simply the function itself That point is evidently a pole of order 3 2 a The singular point is z 0 Since h 2 4 6 3 1 1Zz ilg z z 2 4 6 when 0ltlzlltoo we have m1 and B i 1 101 b Here the singular point is also 2 0 Since 2 2 33 4 4 s s 1 ex4p221711322 22 22 22 z z 1 2 3 4 5 2 1 22 1 231 quotIi3927quot3739Z ZTETZquot39 23 4 when Oltlzllt co we have m3 and B c The singular point of xpfzz is z 1 The Taylor series z 2 2 3 3 exp2ze2z1e2e212zl39 1 2 2239 1 2 z339 1 g W m enables us to write the Laurent series exp22e 1 2 1 22 22 1I co ac 12 z l2 11 z l 2231Z DJquot NZ lt 22 2 Thus m2and Be 1392e Since f is analytic at zo it has a Taylor series representation fZfZoIzzolz zoz Iz zolltRo Let g be de ned by means of the equation 82 fZ Z 0 102 a Suppose that f zo 5 0 Then 1 Z Zo f zo 1 f zo 2 8ltZD fltZogt zzo zzo2 fZof zOf 10 zz0 1 2 z zo 0ltlzz0ltR This shows that g has a simple pole at z0 with residue f zo b Suppose on the other hand that fzoi 0 Then 1 fzo f z0 2 8z z z 1 z zo 2 zzo L 1Iz z0 l 0ltlzzolltRo Since the principal part of g at Z0 is just 0 the point z 0 is a removable singular point of g 4 Write the function 2 f z zz a2 a gt 0 m z 8a3z2 f z z 03 where z Z ans Since the only singularity of z is at z ai z has a Taylor series representation Mai quotai 1 2 z ai2 lzailt 2a W ai z ai about 2 ai Thus 1 z ai f 2 3 ai Iifizaiz an2 o ltlz aillt 2a Now straightforward differentiation reveals that l6a iz 8a3z2 161 z2 4aiz a2 0 z 2 004 and P z z 6105 103 Consequently mm a2i ai and ai i This enables us to write l 2 i 2 fz Zai3 a z 2z at 2z at 0ltzazllt2a The principal part of fat the point z ai is then i2 a2 azi Z ai zai2 z aifi SECTION 67 1 a The function fz 2 22 2 z 1 has an isolated singular point at 2 1 Writing f z z where z 22 2 and observing that z is analytic and nonzero at 2 1 we see that z 1 is a pole of order m 1 and that the residue there is B 61 3 b If we write z 3 211PM where ltz58 1 we see that z 2 IS a Singular pomt of f Since z lS analytic and nonzero at that fz point f has a pole of order m 3 there The residue is hwy 2 16 c The function exp z exp 2 z27r2 39zzrizm39 has poles of order m 1 at the two points z im The residue at z m is and the one at z 7ri is 104 l4 2 a Write the function fz 21 lzlgt 0 0 lt argz lt 27 as l where zzme lulzlgt00ltargzlt27 Q fz z1 The function z is analytic throughout its domain of definition indicated in the y Branch cu X gure below bu O Also i Q N D D1 e logl imam m4 7 1t I e e COS lSm 4 4 43 This shows that the function f has a pole of order m 1 at z 1 the residue there being B 1 Logz b W teth f u n e unc on fz 22l2 z Logz f z if Where 1 cm From this it is clear that fz has a pole of order m 2 at z i Straightforward differentiation then reveals that Logz 7139 2i s eiuzn2 0 8 105 c Write the function 12 00ltargzlt27t as 412 z 2 z whe e z i2 r NZ zi2 Since I ziZl2 4zll2 4 Z 2z i3 and 1 i A 1 i llI2eml4 12 m14 xE IE 1 e 45 5 in l i Res 39 22 02 4 0 M 3 a We wish to evaluate the integral I 3z3 2 z c z 1z2 9 where C is the circle lz 2 2 taken in the counterclockwise direction That circle and the singularities z l 1 3i of the integrand are shown in the gure just below Observe that the point 2 1 which is the only singularity inside C is a simple pole of the integrand and that 3232 3z 2 1 z1z 1zz9 z29 Fl 239 According to the residue theorem then 3232 1 JCz lz29dz 2m m39 106 b 4 a integral Let us redo part a when C is changed to be the positively oriented circle Izl 4 shown in the gure below In this case all three singularities z 1 31 of the integrand are interior to C We already know from part a that 3a3 2 1 Res T z1 z lz 9 2 It is moreover straightforward to show that Res 3z32 3z32 1549i z3i z 1z2 9 Z quot39 1X2 3i z3i 12 and es 3z32 3z32 15 49i z 3i z 1z2 9 Z 1X2 quot 3i 3i 12 The residue theorem now tells us that I 3z32 c z 1z2 9 dz2ml154911549i6m 2 12 12 Let C denote the positively oriented circle Izl 2 and note that the integrand of the Li has Singl a ties at Z 0 and z 4 See the gure below b 107 To find the residue of the integrand at z 0 we recall the expansion 1 fl quot 1 z r lzllt 1 andwrite 1 L 1 1 z Hquot quot3 23z4 4z 1z4 4z3 4 g 4 Z 0ltzlt4 1 Now the coef c1ent of here occurs when n 2 and we see that 2 Res 3 z z 4 64 Consequently J dz Cz3z4 64 3239 Let us replace the path C in part a by the positively oriented circle Iz 2 3 centered at 2 and with radius 3 It is shown below We already know from part a that 1 1 Res z z z4 64 To nd the residue at 4 we write 1 3 1 NZ z z 4 z 4 Where 1 39 z This tells us that z 4 is a simple pole of the integrand and that the residue there is 4 l 64 Consequently dz 1 l 2m 0 Icz z4 64 64 108 5 Let us evaluate the integral L 6 coshn39zdz zz2 1 All three isolated singularities z 0i39i of the integrand are interior to C The desired residues are where C is the positively oriented circle lzl 2 cosh m 2 20 zz cosh 7172 2 1 Z 10 cosh TEZ cosh 722 l zi zz2 1 ZZ z 2 and cosh Irz cosh z CS 2 z 22 ZZ 1 z NIP Consequently I 0081de 27ri111 4m C zz 1 2 2 In each part of this problem C denotes the positively oriented circle lzl 3 a It is straightforward to show that 3 zz2 3z22 1f1 z2 z zlz25z39 f War 1x323 This function 17 f has a simple pole at z 0 and z z M i 1 2 ICZ212z5dz Zm gs zzfz 2m 2 9751 b Likewise z3l 32 1 z1 zz i 1z3 ffm 812 zz1z 2gt39 The function i2 f has a simple pole at z 0 and we nd here that z 2 dz 2m agosLiz 2m 3m J39 231 32 c 1z1 2z 109 c Finally 3 ll z e 1 1 e f 1z3 zzf z z2123 The point z 0 is a pole of order 2 of i2 f The residue is 0 where z z 2 e Z Since 1 232 e 322 UEY 39z the value of O is 1 So 31 ze 1 l L 1z3 dz 2m lifosl f 2m1 2m SECTION 69 1 a Write cscz 39 where pz 1 and qz sin 2 smz qz Since P01 0 q0sin00 and q 0cos01 0 z 0 must be a simple pole of cscz with residue 1 q 0 1 b From Exercise 2 Sec 61 we know that 1 1 1 l s cscz Zglz32 5 0ltIzllt 71 Since the coef cient of 1 here is 1 it follows that z O is a simple pole of cscz the z residue being 1 110 2 1 Write z sinhz zl where 2 zzsinhz 12 pz z smhz and qz z smhz Since 1775139 m39 at 0 q71 39i 0 and q39 m n2 at 0 it follows that sinh ni 39 39 RESET ibgf f 1 1quot w z smhz q m 7 7 b Write 331 32 Wh re Pz expzt and qz sinhz It is easy to see that es expzt pni Res 6Xpzt p7ri m d t sinhz q39en39 exp a sinhz q39m39 ex quot Evidently then Reg expzt I ex39pz t 2 expzmexptmf quotzcos M F smhz F39m smhz 2 3 a Write f Z p zl where pz z and 12 cos 2 42 Observe that 7 lbHm 0 n 01Jr2 Also for the stated values of n P rm mt 0 and qr12r 117 sin mt lnl 0 111 So the function f z z has poles of order m 1 at each of the points cosz 7 zI 2 mr n0i1i2 The corresponding residues are b Write tanhz where pz sinhz and qz coshz q 2 Both p and q are entire and the zeros of q are Sec 34 z nni n 0i1i2 In addition to the fact that mrji 0 we see that nItJi sinh 72 i nm 2 icosmr i1quot 7 0 and 439 1271139 sinh i i 1quot 6 0 So the points z mri n 0ili2 are poles of order m 1 of tanhz the P mzi 1 B 43m uw residue in each case being 112 4 Let C be the positively oriented circle lzl 2 shown just below a b To evaluate the integral Jctanz dz we write the integrand as 172 tanz where pz sinz and qz cosz qz and recall that the zeros of cosz are z 12rn7r n 0i1i2 Only two of those zeros namely z 1717 2 are interior to C and they are the isolated singularities of tanz interior to C Observe that quot 2 1 and Res tanz P 2 1 z7r2 q yr2 PM 139 772 Hence Jctanzdz 2m 1 1 47ri The pmblem here is to evaluate the integral I dz To do this we write the csmhzz integrand as 1 172 39 h 1 sinhZz qz w m P and qz smh2z Now sinh 22 0 when 22 nn i n 0ili2 or when mn39 z 2 n Oi1i2 Three of these zeros of sinh 2z namely 0 and 2 are inside C and are the isolated singularities of the integrand that need to be considered here It is straightforward to show that e 1 amp1l zo sinh2z q 0 ZcoshO 2 113 S 1 p71 i2 1 1 1 zm392 sinh 22 q m392 2 coshn39i 2 cos 71 2 and Re 1 pm2 1 1 1 zmi281nh22 q 7112 2cosh 7ti 2cos 7t 339 I dz 41435 Csinh2z 2 2 2 39 Within C the function 281 has isolated singularities at z mz z0 and zi39n7r n12N To find the residue at z 0 we recall the Laurent series for cscz that was found in Exercise 2 Sec 61 and write csczi l1 z 1 1 3 zzsinz z2 z2 z 3 312 51z 1 11 1 1 0lt kit 23 6 z 302 51z z 114 This tells us that has a pole of order 3 at z O and that z2 sin 2 Res 2 z0 z smz 1 6 As for the points z inn n 12 N write 1 2 h l 2 S W ere PZ and qz z smz 42 Since pin7r 1 0 qimr 0 and q in7t nz z cosnn lquotn27r2 0 it follows that 1 1quot 1quot Res mzzsinz me 1quot n27r2 So by the residue theorem J dz C z2 sinz dz 2m 2 1quot l n272 Rewriting this equation in the form dz quot1 n2 12 4139 Cuzzsinz i 1nl 2 7t and recalling from Exercise 7 Sec 41 that the value of the integral here tends to zero as N tends to in nity we arrive at the desired summation formula lnl 35 quot1 712 1239 The path C here is the positively oriented boundary of the rectangle with vertices at the points i2 and 1392 i The problem is to evaluate the integral J dz Cz2 12 339 115 The isolated singularities of the integrand are the zeros of the polynomial qz z2 12 3 Setting this polynomial equal to zero and solving for z2 we nd that any zero z of qz has the property 22 1115139 It is straightforward to nd the two square roots of 1x i and also the two square roots of l Vgi These are the four zeros of qz Only two of those zeros 45H i zo e 6T and quotzoquotEe 6Ty lie inside C They are shown in the gure below 2i y C 2i X 50 9X L V 2 0 2 x To nd the residues at 20 and 20 we write the integrand of the integral to be evaluated as 2717513 Where pz 1 and 42 z2 12 3 This polynomial qz is of course the same qz as above hence qzo 0 Note too that p and q are analytic at z0 and that pzo at 0 Finally it is straightforward to show that q39Z 4zz2 1 and hence that q39zo 42023 I 2x5 645i 2 0 We may conclude then that 20 is a simple pole of the integrand with residue PZo 1 q z0 243 64239 Similar results are to be found at the singular point Eo To be specific it is easy to see that q zo 20 q39zo NE 6J2 a o the residue of the integrand at 20 being pea 1 11 20 WhoZquot 116 Finally by the residue theorem dz 7 1 1 Jazz 12 3 2 39 2466v i 248mmquot 24539 We are given that f z 1 qz2 where q is analytic at 20 qzo 0 and q zo 0 These conditions on q tell us that q has a zero of order m 1 at zo Hence qz z zogz where g is a function that is analytic and nonzero at 20 and this enables us to write W 1 M 2 20 Where 2 gz1239 So f has a pole of order 2 at zo and 2g z give zo gzO 3 But since qz z zo gz we know that q z z Zog z 82 and q Z z Zo g z 23 zv Then by setting 2 20 in these last two equations we nd that q zo g20 and q zO 28 Consequently our expression for the residue of f at 20 can be put in the desired form 13308 39 a To find the residue of the function csc2 2 at z O we write 2 csc z where qz sinz 1 qz2 Since q is entire q0 0 and q 0 lab 0 the result in Exercise 7 tells us that q 0 Res csczz 0 Z 13903 117 b The residue of the function zz2 2 at z0 can be obtained by writing 1 1 h 2 28 qz2 w m 42 2 Inasmuch asq is entire q00 and q 0 1 at 0 we know from Exercise 7 that Res 1 q 0 zo zz22 q 03 39 118 Chapter 7 SECTION 72 l 1 around the simple 1 To evaluate the integral we integrate the function f z 2 0 x 1 2 closed contour shown below where R gt 1 3 CR 0 0 r E x Weseethat R J f quot J z 2m39a Rx 1 52 1 where BRes 21 ReS 39 l in zt z 1 139 Z lXZ39i l Zl zi 2 Thus R dx 4 JT 39J x l sz 1 R Now if z is a point on CR lz2 1lgtlzl2 1 R2 l and so ale 7tR s gt0 as Rm 2 c R 1 11 R2 dz ICRZZI Finally then G dx 0 0 119 39 dx 1 The mtegral can be evaluated usm the function z and the same 39I 1 2 12 g f 22 12 simple closed contour as in Exercise 1 Here R dx dz Lx2 J CR ZZ where B Eels Since 1 z 1 h zz 1 2 igt2 w cm 2 20 we readily find that B i and so 1 f4x 39 dz I Rx212 2 Cz21239 If 2 is a point on CR we know from Exercise 1 that z2l2R2 1 thus 7139 dz IrR F S co Hzle R24 1 290 as R gt 1 7 The desired result is then FLA ILJE x212 2 o212 439 We begin the evaluation of by finding the zeros of the polynomial 2 1 which are 0 the fourth roots of 1 and noting that two of them are below the real axis In fact if we consider the simple closed contour shown below where R gt1 that contour encloses only the two roots 1 i zem4 d I If 2 an 1 i 1 i e13a4 em4enzl2 22 2 5 45 45 120 y CR X X z2 Zr 0 I x Now R dx dz Lm qm 2quot le where 1 Bl Izensz l and 32 13Ezsz4l39 1 The method of Theorem 2 1n Sec 69 tells us that z1 and z2 are ample poles of 4 1 and z that 13135511 and 32 13z2z2 421 21 4 422 z2 4 since 2 1 and z 1 Furthermore l lLL 139i 53quot 4 4J392 2 2 N3quot Hence fih f 51 Rx4l 15 sz4l Since dz 77R S gt0 R gtoo JCRZ41 R41 as wehavc jig or jdxn x41 J 0x 1 z 39 eo 121 quot 2 dx 4 We Wlsh to evaluate the integral L We use the simple closed contour 0 x x2 1x2 4 shown below where R gt 2 2139 i l l w We must find the residues of the function 2 z t 39ts 39m l l f 221zz4 a 1 51 pe poes ziandz2i Theyare Bl Resfz Z2 i quot 2in2 4 m 6i and 32 RCSfz z l z2i Z2 2quot Thus R dex I zz dz L062 1x2 1 ICR zz 1ZZ 32 or ii 2 I zzdz RJC2 lx2 4 3 cl 22 1Xz2 4 If 2 is a point on CR then Iz2 1l2lzI2 1 R2 1 and Iz2 4l2lzl24I 12 4 Consequently n J ZZdz 395 R3 39 5 gt0 R gt c z2 1z2 4 R2 1R2 4 1 1 1 4 as 39 F 77 and we may conclude that quot xzdx 7r 39 x dx 7r JT r or J T T39 x lx 4 3 ox 1x 4 6 122 x2 x2 9x2 42 fz 5 The integral f can be evaluated with the aid of the function 0 2 z z2 9z2 42 and the simple closed contour shown below where R gt 3 y CR X 3i X 2i 0 t E x We start by writing I 2 2 x dx z dz 2 B B Lx29x2 42 Jcu 22 9z242 m 1 2 where z2 z2 R d B R Bl zesis zl an 2 22 Now z2 3 31 39 2 7 z 3zz 4 23 501 To nd Bz we write 22 252 z2 h z2 9z2 42 z zi2 w cm W 22 9z2i2 Then 32 2i 5 This tells us that fxzdx L J 2 Rx2 9x2 42 100 c 22 9z2 4239 Finally since gt0 as R gtoo I z2 dz I lt nR c 22 9x22 4 R2 9R2 42 we find that xzdx 7r quot xzdx 7t J a T zquotquotquot 0 1 2 5 7 39 x 9x 4 100 0x 9x 4 200 oo 123 7 In order to show that I xdx 7 39 x21x22x2 5 D o we introduce the function fZ z2 1z2 Zz 2 and the simple closed contour shown below Observe that the singularities of fz are at i 20 1i and their conjugates i 20 1 i in the lower half plane Also if R gt 42 we see that I fxdx C fzdz 271130 3 3 R where l 3 BResz 5 o Hf 22lz20z20 10 1039 and l l Res z Z 39 Bl ziz22z2z 10 539 Evidently then f xdx 7139J zdz Rx21x22x2 5 Cnzzlz22z239 Since J zdz J zdz 71R2 O Cnzzlz2222 CnzzlzzozZol R21R 2 as R gt no this means that fx x 2 a x2 1x2 2x 2 5 39 This is the desired result 124 dx 271 8 The problem here is to estabhsh the integration formula using the simple 1 x3 1 3 closed contour shown below where R gt 1 y Rena3 y 0 q R x 1 3 namely z0 e39quot that IS interior 1 1 to the closed contour when R gt 1 According to the residue theorem dz dz dz 1 L ZS39H Ickz3l Iczzs1 2711620823 1 There is only one singularity of the function f z where the legs of the closed contour are as indicated in the figure Since C1 has parametric representation 2 r O S r S R and since C2 can be represented by z ream 0 S r s R d d R iznlsd i R d lag Z L z quote quotquote r z31 z31 re 2quot 3 1 Furthermore I 1 1 Res zzo Z3 1 332339 36mm Consequently R dr 2m dz 1 emu3 r3 3e12 3 JC Z3 But dz 1 27rR s gt R gt oo an n 123 1 3 033 This gives us the desired result with the variable of integration r instead of x f dr 2m 27ri 7c 31 r31 3ei27r3 ei47t3e i61r3 3ei27r3 e i27r3 quot39 38in2 3 s 39 0 125 9 Let m and n be integers where 0 S m lt n The problem here is to derive the integration formula quot 2m Lin quotix 1csc2m17r o x 1 2n 2n a The zeros of the polynomial 22quot 1 occur when 22quot 1 Since 1 2quot 6Xpi k 0 12 2n 1 it is clear that the zeros of 21quot 1 in the upper half plane are ck expim k012n1 2n and that there are none on the real axis 1 With the aid of Theorem 2 in Sec 69 we find that Zlm C2 1 2 m quotH l k012 n1 m zzquot 1 2ncfquot l 2n k 1 n we can wnte 2 Putting a m Clio n1 expi 2k 1 n 2n 1 expl 2k l2m 1 2n exPi2k 1 7 ei2k1a 2m 1 Res mm k0l2n 1 zc zzn 1 Zne In view of the identity see Exercise 10 Sec 7 111 2amp1quot zeal co 1 13 126 then quot 1 2m n l ilan ia i2an Z m in i201 k m gal e e m e l 2712 RCSW 72e I e e e l amp 390 k0 mt Z n 1 2 e n e39 e ei2m11r 1 n n e39 equot n 2quot equotquotquot nsina 39 c Consider the path shown below where R gt 1 The residue theorem tells us that 2 211 R x z2m nl z dxJ dz 2n39i Res ix 1 Cu zz 1 i6 zzquot 1 2m 2m x 71 2 dx ix 1 nsina Ln zzquot 1dz Observe that if z is a point on CR then lzzmlRzm and lz2quot12R2quot1 Consequently and the desired integration formula follows 10 The problem here is to evaluate the integral I dx 0 x2 a2 12 39 where a is any real number We do this by following the steps below 127 a Let us first nd the four zeros of the polynomial b qz z2 a2 1 Solving the equation qz 0 for 22 we obtain 22 aii Thus two of the zeros are the square roots of a i and the other two are the square roots of a 139 By Exercise 5 Sec 9 the two square roots of a i are the numbers 1 zo 7 5 VAa l VA a and 20 where A Va2 1 Sincei39zo2 z ai a i the two square roots of ai are evidently zo 2 The four zeros of qz just obtained are located in the plane in the gure below which tells us that 250 and 20 lie above the real axis and that the other two zeros lie below it y 0 o 20 zo 0 x o o 39zo 2 Let qz denote the polynomial in part a and de ne the function 1 f 39 qz which becomes the integrand in the integral to be evaluated when 2 x The method developed in Exercise 7 Sec 69 reveals that z0 is a pole of order 2 of f To be speci c we note that q is entire and recall from part a that qzo 0 Furthermore q39z4zz2 a and z ai as pointed out above in part a Consequently q zo 420 z a 4izo 0 The exercise just mentioned together with the relations z a i and 1 a2 A2 also enables us to write the residue B1 of f at 20 B q zo 12z3 4a 3zga 3ai a a i a i2a23 q zo 1 423 16iz z0 16iaizo a i 1642 As for the point 20 we observe that crew m and q 2Z39Tzquot 128 6 Since qZo O and q Zo q z 4i0 at O the point 20 is also a pole of order 2 of f Moreover if B2 denotes the residue there 2 q zo q zo q zo q Flto3 q zg3 4 zo3 I 1312 15r1El 2iImBl 2 12Im az2a 3 8A1 zo We now integrate f 2 around the simple closed path in the gure below where R gtz0 and CR denotes the semicircular portion of the path The residue theorem tells us that R fxdx fol fzdz 271quot191 32 f dx Im ai2a23 4 dz Jc2a212 4A2 Zo Cqz239 R In order to show that dx 139 RagLu qz2 we start with the observation that the polynomial qz can be factored into the form 42zZozZozEozfo 39R Recall now that R gt Izol If z is a point39on CR so that z R then lziZOIZIIzIIzOH R lzol and IziZolzl lzl IEOI I R Izol 129 This enables us to see that qzl 2 Rlzol4 when 2 is on CR Thus l l 2 S 3 42 Rlzol for such points and we arrive at the inequality 7t 1 75R quot 7 dz 5 R In qz2 Rlzol8 I 301 R which tells us that the value of this integral does indeed tend to 0 as R tends to co Consequently quot dx 2 a i2a2 3 P Im 3 x2 a2 12 4A2 l zo But the integrand here is even and 1m ai2a2 3 Im V z ai2a2 3 VAa i1A a 2 7T39 39E IN i 39 7 Fr wT a 39 So the desired result is dx Ix2 a212 84125143 M 3 A quotA 0 where A Va2 1 SECTION 74 1 The problem here is to evaluate the integral I this we introduce the function f z cosxdx x2 a2 x2 b2 1 z2 a2z2 b2 where agtbgt0 To do whose singularities ai and bi lie inside the simple closed contour shown below where R gt a The other singularities are of course in the lower half plane 130 According to the residue theorem R e dx iz Luz a2x2 b2 21quot we dz quot mm 32 where B Resfze ei quot 6 1 zai z aiz2 72 my 2ab2 a2i and B Res f zei eiz b 2 wt z2 a2z bi 2 2ba2 b2i39 That is R e dx 39 6 9 139 LG azx2 b2 a a2 b2 5 a f2e dz or I cosxdx 7r e39b equot Rx2a2x2b2 az bzk b a Re I f ze dz CR Now if z is a point on CR MR WhCI39C Mkm and leizl e y S 1 Hence iz i 713R IRCJCR fze dz S J C39 fZ zdzl S MRER m 0 38 R 00 So it follows that cosxdx 7 f e agtbgt0 I 4 x2a2x2b2 a2 b2 b a quot cosax 2 This problem is to evaluate the integral I 21 dx where aZO The function x 0 1 fzzz1 contour shown below where R gt1 has the singularities ii and so we may integrate around the simple closed 139 gt11 131 We start with R in e ixz 1dx ifze39 dz 2213 where B Resfzeiquot 1 e Fquot 2 i m 2i Hence R em i id quote 39 aimquot zdz or R cosax L x2 1 dx quot3 quotRei zkmdz Since IfzlS MR where M 2 R 1 we know that 75R Re z emdz S 2 emdz S in f RM and so I cozsax dx ne39 1 1 That is j 3de ea a20 o x 1 2 To evaluate the integral J 212 dx we rst introduce the function x 0 fz z z z23z zlzZl where 21 312 The point zl lies above the x axis and 21 lies below it If we write z where z M Z Z Z El fzei22 132 we see that 21 is a simple pole of the function f zei2z and that the corresponding residue is iexpGZxg exp2x 2J3 2 39 B1 11 Now consider the simple closed contour shown in the gure below where R gt 5 3 CR zlt J55 ot x Integrating f zequot22 around the closed contour we have R xei2x 39 22 Lx23dx sz Iq z dz39 Thus R jxsmx w x2 3 dx Im2m39Bl ImJ CR fzem dz Now when 2 is a point on CR fzS MR where MR O as R oo R2 3 and so by limit 1 Sec 74 39 ilz girlie fze dz 0 Consequently since S lImJ39CK fzenz dz jg fzequot2z dz we arrive at the result xsinx quotxsinx dx ex 2w or p 2 x2 3 Jx23 dx gexp 243 133 no 3 a 6 The integral to be evaluated is I x fmax dx where agt0 We define the function x 4 3 fz 4 4 and by computing the fourth roots of 4 we nd that the singularities z zl J ieu ul4 1i and 22 ei3m4 Wear4ng LN1 1i both lie inside the simple closed contour shown below where Rgt The other two singularities lie below the real axis y CR X X 22 Z1 0 1 x The residue theorem and the method of Theorem 2 in Sec 69 for nding residues at simple poles tell us that x e in immigme dzzmaa 32 where B z eta 2138mm e142 eralr eaem 221 2 4 4213 4 4 4 and B Z eiaz zzaeiazz eiaz end 10 eae m 2 Fl 6 4 4z23 4 4 4 Since equot Ie39i39z 27MBl 32 me m e cosa we are now able to write R 3 x srnax in IL x44 dx n e cosa ImIC fze dz 134 Furthermore if z is a point on CR then 3 R R44 fzlSMR where MR and this means that gt0as R 9oo aOuRam Im on fze dzls LR 162de according to limit 1 Sec 74 Finally then 3 x smax dxarequotcosa agt0 J x44 on 3 x Slnxdx 8 In order to evaluate the integral I r T we introduce here the functlon 0 x 1x 9 Z3 z 22 1z2 9 39 consider the simple closed contour shown below where R gt 3 Its singularities in the upper half plane are i and 3i and we x Since R ix z3eiz 1 235 f 2 1zizz9 Ff E and iz za ii 9 13 fze 1 22 1z3i 166339 the residue theorem tells us that R x3ei dx 2 1 9 I x21x29 16 f 28 dx 2ml 6l 6e39 R R 3 x smxdx 7r 9 it I x21x29 E39l lmicfze dz R 135 Now if z is a point on CR then R R21R29 as R gtoo fzlSMR where MR So in view of limit 1 Sec 74 iImIC fzequotzdz s gt0 as R m Ln f ze dz and this means that I xsinxdx l91 or x3sinxdx l x21x29 8e e2 x2lx29l6e e2 39 0 sinxdx 2 can be found with the aid of the x 4x 5 The Cauchy principal value of the integral I 1 22 4z 5 Using the quadratic formula to solve the equation z2 4z50 we find that f has function f z and the simple closed contour shown below where R xg singularities at the points 21 2 H and 21 2 i Thus f z IT where 21 z zizzi is interior to the closed contour and 21 is below the real axis y Cl Xz The residue theorem tells us that R ei dx IZ ix24x5 LlfZe dz 2mB where ll 239 BRes e e zzl z z1z zl zlzl andso R sinxdx 27thizl IT quotIm x 4x5 zl zl Im LR fze dz R 136 R smxdx 7 I 2 1 sz 4x5 e sm Imjm zk dz39 Now if z is a point on CR then leizl equoty 1 and lfzS MR where MR RlzdRle R 52 Hence i iz 7rR IIITIIC fZe dz 5 IcnfZe dZIS MR R m o as R and we may conclude that PVIu smxdx sin2 39 x24x5 39 x 1cosx 2 dx we shall use x 4x5 10 To nd the Cauchy principal value of the improper integral I z1 z1 zz4z5 z zxz Zl same simple closed contour as in Exercise 9 In this case the function fz where z12i and El 2 1 and the R x1ei dx 39 d 2 393 Rx24x5 Jcfze z m where it iz 2i BRes z1e z11e 1le i1 ZZlZZl zzl 261 Thus R x1cosx dxR 2 B 1 Rx24x5 e 7 Ick zk or f x1cosx 7t x2 4x5 dx st cos2 J Ckfze dz R Finally we observe that if z is a point on CR then R1 Rl Rlz XRlfll 1243 lfzlltMR where MR OasR oo 137 Limit 1 Sec 74 then tells us that lReL fUeidelS L fzei dz gt0 as R gtoo and so RV x1cosx 7 dx I 2 x2 4x 5 e 51quot 0 52 12 a Since the function f z expizz is entire the CauchyGoursat theorem tells us that its integral around the positively oriented boundary of the sector 0 S r S R 0 S 9 S 7174 has value zero The closed path is shown below A parametric representation of the horizontal line segment from the origin to the point R is z x 0 S x S R and a representation for the segment from the origin to the point Re quot is z rem O S r S R Thus R R a 2 Ie dxIc equotzdze quot 4Iequotzdr0 o quot o R R ix iu4 r i 1 Ie dx e Ie dr J39Cle dz 0 0 By equating real parts and then imaginary parts on each side of this last equation we see that R i 1 R 2 rz i 1 coax dx ge dr Rej clez dz and Isinx2dx Vl Z Iequotzdr ImL ei zdz 138 b A parametric representation for the arc CR is z Re 0 S 6 S 75 4 Hence 1174 e L eizzdz I 4 2 10 9 z 1 39R Rte d0 zR Je R New co 669476 0 0 Since lei 529 1 and leiel 1 it follows that c ehzdzl s R Te R d9 quot 0 Then by making the substitution 26 in this last integral and referring to the form 3 Sec 74 of Jordan39s inequality we nd that iz Ruiz R7 sin R 71 71 j e dzs je d s gt0as R m Cu 2 o 2 2R2 4R c In view of the result in part b and the integration formula Iequotzdx i2 75 it follows from the last two equations in part a that 2 1 7r 1 7t c0305 dx and smx2dx SECTION 77 1 The main problem here is to derive the integration formula Jcosax 20080375 dx 3251 a a 2 0 b 2 0 0 X using the indented contour shown below y 139 Applying the CauchyGoursat theorem to the function e39 z equot z fz 12 we have L fzdz L fzdz L fzdz L fzdz 0 L fzdz L fzdz Lfltzdz L mm Since 11 and L1 have parametric representations Azrei0 rpSrSR and Lzrequot39 rprR we can see that ibr L fzdz L fzdz L fzdz L11 fzdz 52 drIe39iquotr e dr R iar iar ibr ibr R J e e 2e e dr 2J cosar 2cosbr dquot r P r p R le dr L fzdz L zwz In order to nd the limit of the rst integral on the right here as p O we write 1 iaz iaz2 iaz3 ibz isz ibz f 391ETTTquot39quot11TTT39 i 39 lm 0ltlzlltoo From this We See that z 0 is 35911219991Q9fVfZyWiElTrCSidtle ia 17 Thus 21 C iit lcfltzdz Bomiltabmnut b x m mim MV 140 As for the limit of the value of the second integral as R 00 we note that if z is a point on CR then le llequotquotl e e39b 11 2 z s s IzI2 R2 R2 R2 Consequently 2 27 chfzleSF R R gtO as Roo It is now clear that letting p gt 0 and R gt oo yields ZJ cosar cosbr dr b a 0 r This is the desired integration formula with the variable of integration r instead of x Observe that when a 0 and b 2 that result becomes 2 J4 003 x dx m o x But cos2x 1 28in2 x and we arrive at 2 sm x n j i dx o x 2 2 Let us derive the integration formula l a7t x 1 3 x212 4cosa7t2 M where x I expaln x when x gt 0 We shall integrate the function z expa log z 7 371 2212 quot 2212 39239 O 3ltargzlt 2 fz whose branch out is the origin and the negative imaginary axis around the simple closed path shown below 141 Branch cut By Cauchy39s residue theorem L fzdz Cl fzdz 1 fzdz CF fzdz 27d 133 s fz That is L mm L fzdz 2m 133s m JG mm jg mm Since lezrei rpSrSR and Lzzrequot rpSrSR the left hand side of this last equation can be written R Kurd1390 R eulnri e Ur Iquzdz Jlafzd239 r2l2 dquot r2n2 3 dr d R a R a R r r r dquotquot d1 quotIr ma r2l2 r e rZH 39 H r212 Also Res i where a fz z lt z 2 02 the point z i being a pole of order 2 of the function f z Straightforward differentiation reveals that Iz eallogzlaz 22 z i3 142 and from this it follows that Res f z dew22 zx 4 We now have but R r quot1 0 ran2 1 6 dr 2 e JG fZdZ ICR Once we show that E Icpf2dz0 and chfzdz0 we arrive at the desired result I r dr la a e39fa z 7rla 2 1 an o r2 12 2 1 e39 quot e39quot quot 2 4 emquot2 e39mZ 4cosa7r 2 The rst of the above limits is shown by writing paw 1 p2 2 and noting that the last term tends to 0 as p 0 since a l gt 0 As for the second limit 5 LP fz dz up Pa 1 p22 l RaH F iii1amp3 s 7rR Rz lf 1221 L 1 2 R4 an fzdz and the last term here tends to 0 as R gt oo since 3 a gt O 3 The problem here is to derive the integration formulas quotVilnx 72 11 2 and 12J 2 d 7l7 o x 1 6 0 x 1 13 by integrating the function z13 logz e13logz logz llgt0 75ltar zlt z2l 221 z 2 g 2 fz 143 around the contour shown in Exercise 2 As was the case in that exercise L fzdz J11 fzdz 27ril gsfz C fzdz fc fzdz Since eI3logz 10 g z fz z where ltzgt 2 1 21 the point z i is a simple pole of f z with residue 71quot irr6 Resfz z e 21 4 The parametric representations I1zzrei rpSrSR and llzre quot rperR can be used to write RWlnr mnRWInanRF L1 r21 dr and IL fZdze Thus kWhquot mn i nanW 72 W6 1 2 1 dre r2 1 drte J CpfzdzLR fzdz By equating real parts on each side of this equation we have RWlnr R Wlnr R V 72 d 39 ii r21 rcosn3 r2l dr nsrn7r3 r2ldr 2 smut6 Re J f zdz ReJC f zdz and equating imaginary parts yields V lnr r r21 R R W n2 smn 3J dr rtcosn39 3J2 dr cosn 6 p pr 1 2 Im IO fzdz Im Ll fzd2 3 1 Now smut3kg cos7r32 sm7r6 cos7t6if and it is routine to show that in CF fz dz 0 and 13 q fzdz 0 144 Thus 2 Eijmrdr J 37 d 20 r 1 2 or 1 4 2 3EJ Vflmdr if dr d 2 0 r 1 Zor 1 4 Thatis 2 1135 2 2 4 2 11 12m5 2 2 4 Solving these simultaneous equations for Ill and 2 we arrive at the desired integration formulas 4 Let us use the function 2522 375 fZ zzl lzlgt0 2ltargzlt2 and the contour in Exercise 2 to show that an 2 3 u lrzix dx and J lznx o x 1 8 o x 1 dx0 Integrating f z around the closed path shown in Exercise 2 we have L fzdz L fzdz Zm39IEgS z JG fZdz Jen fzdz Since mpg where ltzgtM Zl zl the point z i is a simple pole of f z and the residue is 2 Rss z 09 1121L2i i H l 2 81 Also the parametric representations lezrei rpSrSR and llzreiquot rpSrSR 145 enable us to write R 2 R 2 911 M jammy r21dr and Joya i r2 dr Since R 2 R R lnr 2 dr lnr 2 qufzdZJquzdz 2 r21dr lrzl mgr2ldr then R 2 R R lnr 2 dr lnr n393 zlmd r 2 2 quotlr21d 4 LWW Lf z dz39 Equating real parts on each side of this equation we have Rlnr2 R dr n3 2 a 2 r21dr n r2l 4 ReJCpfzdz ReLRfzdz and equating imaginary parts yields 271E lnr drImJC fzdz ImIC fzdz 2 p r 1 It is straightforward to show that 135 La fz dz 0 and 139 CR fzdz 0 Hence or21 4 2 3 21041142 IL 0 r 1 and co ZEJ lzn L dr 0 0 r 1 Finally inasmuch as see Exercise 1 Sec 72 2 39 0r 1 2 Idr 7 we arrive at the desired integration formulas 146 5 Here we evaluate the integral L W air where agtbgt0 We consider the 0 x ax 17 function 1 13 6Xp 10gz zazb zagtzb fZ lzlgt00ltargzlt27c and the simple closed contour shown below which is similar to the one used in Sec 77 The numbers p and R are small and large enough respectively so that the points z a and z b are between the circles Branch cut R x A parametric representation for the upper edge of the branch out from p to R is z rei0 p S r S R and so the value of the integral of f along that edge is If explnri0 drzjg V d r ar b p r ar b r A representation for the lower edge from p to is R is z rem p S r S R Hence the value of the integral off along that edge from R to p is l R exp 1nr12 dr filmIf V p r ar b r ar b According to the residue theorem then fi dr Jfzdzei2nl3f Vdr B p r ar b CR P r ar 1 CF 2 147 where eXP 1 10ga expllna i710 BI Rcsfz 3 3 elf3 V 2 quotd b a b a b and ex 110 b ex llnbiyr B Resfz p 3 g P 3 em393 2 zb ba ba a b Consequently R V Ziaems W it3 g 1 e rarb dr ab fzdz Now 275 dz 5 v 2 gt 0 gt 0 LpfZ a pbp P a pbp as p and W 2 22122 1 ICRfZdZS Rakb RaRb39W 9085R Hence J V dr zmoem333Vb e inS 2mW V395 0 rarb 1ei2x3a b e ms ear3 e i1rl3ab rum Viaquot nVZ VE225VE VE sin7t3a b J 3943 a b39 701 17 Replacing the variable of integration r here by x we have the desired result v gym vs xaxbdx 39 a b agtbgt039 148 6 a Let us first use the branch 1 12 cxp logz z 2 2 273 fz z21 z21 lzlgt0 2ltargzlt 2 and the indented path shown below to evaluate the improper integral I dx OVx2l39 y V R x Branch cut Cauchy s residue theorem tells us that L1 fzdz jg fzdz L fzdz GP fzdz 2m 135s fz a Lfzdz J14 fzdz ZmIESis z Icpfzdz L fzdz Since l1zzrei rpSrSR and L2zreiz rpsrltR we may write J fZdz J fzdz j dr 43 dr 1 if dr L39 Li p 7r21 p502 1 DID WU 1 Thus R dr aquot We 1 2m1333fz lc fzgtdz L fzdz b 149 Now the point z i is evidently a simple pole of f z with residue Res fz 12 exP1 gi exPmlig i z 39 2 2i 2139 2139 45 Furthermore Icpfzdz 3 1735 90 as p60 and LR fzdz 5 WE 7 R21R 1 0 as R gtoc R Finally then we have quot dr 7tl i 1 45 a which is the sameas jdx 1 Echo 4539 39 dx To evaluate the improper integral we now use the branch 502 1 39 ex 110 212 P 2 gz zzlquot Izlgt 00ltargzlt27t fz z21 and the simple closed contour shown in the gure below which is similar to Fig 99 in Sec 77 We stipulate that p lt1 and R gt1 so that the singularities z ii are between CF and CR Branch cut x 150 Since a parametric representation for the upper edge of the branch out from p to R is z re p s r S R the value of the integral of f along that edge is R expln r i0 R 1 1 4 39 d r21 Woun r A representation for the lower edge from p to is R is z re p S r S R and so the value of the integral of f along that edge from R to p is fexp1nri27r dr equotquot R R 1 1 r21 Wr2ldr 7r2ldr p Hence by the residue theorem R 1 R 1 d d 39 15021 T afZ Z I 50Ddr fzdz 2mBlBz where Res z Bl z 2i 2i 2i 1 1 7 12 3XP 31031 CXP Elnl 131 e irr4 zi zi and 1 1 37 zlZ BXPquotE 10g l exp ln 1 eianlt Bz13ltgsfz zi z i 2i 2i 2i That is gimm 7te quotquot an J fzdz J fzdZ c c Since Jcpfzdz s 3 gao as p gt0 and L fZdz39lt 1 WERE 1 Oas R gtoo E 15 1 we now nd that I 1 dr ire inn e i3nl4 irein ei37r4eiz o 4er 1 39 2 2 era4 e m4 COS 2 4 539 When x instead of r is used as the variable of integration here we have the desired result jdx 1 O 1x2 1 2 39 SECTION 78 1 Write T d9 I 1 393 dz 54sin9 C54z z39liz C2z25iz2 0 where C is the positively oriented unit circle z 1 The quadratic formula tells us that the singular points of the integrand on the far right here are 2 i 2 and z 2i The point z i 2 is a simple pole interior to C and the point z 2i is exterior to C Thus 21 d6 1 1 1 271 2mRes 2m 2 39 J 54Si119 z 2222 5iz2 4z5izil2 m3i 3 0 2 To evaluate the de nite integral in question write j W j 1 34 de 1sin29 01zzquot2 iz Cz46z2l 2i where C is the positively oriented unit circle lzl 1 This circle is shown below 152 Solving the equation z2 2 6z21 0 for 22 with the aid of the quadratic formula we nd that the zeros of the polynomial Z4 622 1 are the numbers z such that z2 31 2x5 Those zeros are then 2 ix3 245 and z 113 2J5 The first two of these zeros are exterior to the Circle and the second two are inside of it So the singularities of the integrand in our contour integral are Z1 Zz 39Zl indicated in the gure This means that 1 d6 2 i1sin20 mail32 where 4iz 4iz i i i 1 EES 1 1 zzlz 6z21 4213 1221 zfs 3 2454 245 and 4iz 4iz i i 13 12x38 2 Zznz46zzl 4zf1zz1 4 3 24539 Since 39 27 E 27513 B2m l 27 1 2 f2 5 J2quot the desired result is d9 2 J1sinza f 7 Let C be the positively oriented unit circle zl 1 In view of the binomial formula Sec 3 l2n quot 1quot 1 zz 1 2quotdz 1 s1n2 9d9 sm2quot9d6 L J 2 2 IC 2139 iz 22quot1 1quotiJC z dz 1 n 2quot 2n k 22n11niJC k Z lkz 1dZ 1 n 2 k 2n 2kl 22nl1ni z k Icz dz k0 153 Now each of these last integrals has value zero except when k n L z ldz 2111 Consequently n 1 2n 1quot2m 2n 2 MO 5111 22nl r102 22 n 02 SECTION 80 5 We are given a function f that is analytic inside and on a positively oriented simple closed contour C and we assume that f has no zeros on C Also f has n zeros z k 1 2n inside C where each 2 is of multiplicity m See the gure below y C 0 I x The object here is to show that Zf z quot L z dz ZmEmkzk To do this we consider the kth zero and start with the fact that fz z 2k gz where gz is analytic and nonzero at 2 From this it is straightforward to show that ZfZ me 18 my Z zk mkzk zg z zgiz mkzquot m f z z zk gZ z 2 32 gz z zk Since the term 3 9 here has a Taylor series representation at zk it follows that 51 z z has a simple pole at z and that Res zf z mkzk 2 f z An application of the residue theortm now yields the desired result 154 6 a To determine the number of zeros of the polynomial z6 52 z3 2z inside the circle lzl 1 we write f z 524 and gz 26 z3 2z We then observe that when 2 is on the circle fzl 5 and gzl s IzI6 Izl 2lzl 4 Since I f z gt lgz on the circle and since f 2 has 4 zeros counting multiplicities inside it the theorem in Sec 80 tells is that the sum fz8Z 26 524 23 22 also has four zeros counting multiplicities inside the circle b Let us write the polynomial 2z4 223 222 22 9 as the sum f z gz where fz 9 and gz 22 2z3 2z2 2z Observe that when 2 is on the circle Izl 1 fzl 9 and lgzl s 21zl 2zP 2le 2lzl 8 Since I f z gt gzl on the circle and since f z has no zeros inside it the sum f z gz 2z4 223 222 2z 9 has no zeros there either 7 Let C denote the circle zl 2 a The polynomial z4 323 6 can be written as the sum of the polynomials fz 323 and gz z 6 On C fzl alzP 24 and gzl Iz 6ls lzl46 22 Since I fzlgtlgz on C and fz has 3 zeros counting multiplicities inside C it follows that the original polynomial has 3 zeros counting multiplicities inside C b The polynomial 24 2223 9z2 z l can be written as the sum of the polynomials f259z2 and gzz4 223z 1 On C I fz 9lz2 36 and Igzllz 223 2 1 s lzl42z3lzll 35 155 Since fzgtgzl on C and fz has 2 zeros counting multiplicities inside C it follows that the original polynomial has 2 zeros counting multiplicities inside C c The polynomial zs 323 22 1 can be written as the sum of the polynomials fzzs and gz3z3zzl OnC lfzllz5 32 and lgzl I3z z2 115 3z3 lzl2l 29 Since fzgtlgzl on C and fz has 5 zeros counting multiplicities inside C it follows that the original polynomial has 5 zeros counting multiplicities inside C 10 The problem here is to give an alternative proof of the fact that any polynomial Pz ao axz aizquot azquot a 0 where n 21 has precisely n zeros counting multiplicities Without loss of generality we may take aI 1 since n I an PZ anamp z quotup iznl 51 Let fz Zquot and 82 do alz a1zquotquot Then let R be so large that Rgt1laollaIIa l If z is a point on the circle C lzl R we nd that lgz Slaol lalllzl lan1zlquot l laol lqu laHlR39quot1 ltla0Rquot la1lRquot l alRquot l laollall lalRquot39l lt RRquot391 Rquot lzlquot lfzl Since f z has precisely n zeros counting multiplicities inside C and since R can be made arbitrarily large the desired result follows 156 SECTION 82 1 The singularities of the function 293 s4 4 F s are the fourth roots of 4 They are readily found to he s ew k 0123 r o E Iii E and 45139 See the gure below where y gt 2 and R gt J5 7 fr 7iR 1an u YiR The function 2538 s4 e Fs has simple poles at the points so f2 sl 1Ei s2 w and s3 i and 233Kquot 3 25383 3 l t s4 4 quot2 4s3 22quot 3 2 Rese Fs i Res n0 quot39 n0 5quot 1 t 1 Ni 1 51 l 39w392t e e 39 2 2 2 e ze e45 e Jit eix i ei 2 2 cosh t cos fZ t 157 Suppose now that s is a point on CR and observe that sl7Remls 7RRy and lsllyRe 9 2 Iy Rl R ygtw 2 It follows that 12s3l 2Isl3 s 2R y and Is 4l2ls 42Ry 4gtO Consequently 2R 103 IF ls co S R744 gt0 as R gt This ensures that f t cosh it cos t The polynomials in the denominator of 2s 2 Fm s 1s2 2s 5 have zeros at s 1 and s 1i39 2i Let us then write eStFs e 3 1s slsSl where 31 1 2i The points 1 s1 and 3391 are evidently simple poles of equotFs with the following residues e 2s39 2 BlRese Fs e39 z1 s 31 s s1 ml 1 B2 ResequotFs e 251 2 1 lequotem slum 51 5 R F e3quot2 2 e 2512 l awn B3 lsIe 5 511 1sl mum 1 32 22ee 39 158 It is easy to see that 1 i 1 B1BzBs e 2 2 e2 2 2e e 2 e12 e i23 em e i2 39 e 39e T 2 e 39sm2tcosZt 1 Now lets be any point on the semicircle shown below where y gt 0 and R gt E y 7iR l MR Since lsllyRew57RRy and IsIIyRe Izly RIR ygtJ we ndthat 232l2lsl22R72 ls1l2lsl 1lZR y 1gtO and If 2s5ls sns I 2 Isl Is 2 2R y2 J 2 gt0 Thus l2s 2I lt 2Ry2 s1s2235Ry1Ry2 2 0 as R gt Fs I and we may conclude that f t equotsin 22 cos 2t l 159 4 The function 2 2 s a 173 S2a22 agt0 has singularities at s iai So we consider the simple closed contour shown below where y gt 0 and R gt a y Fr 7 CR aiX h L 0 7 ai X u 7H Upon writing 2 2 Fs 4quot 2 where s T 2 swat sal we see that s is analytic and nonzero at so ai Hence s0 is a pole of order m 2 of Fs Furthermore F s F E at points where F s is analytic Consequently 50 is also a pole of order 2 of F s and we know from expression 2 Sec 82 that Res equot F s Res e Fs 2 Ree 171 7229 where b1 and b2 are the coef cients in the principal part bl 2 s at s at of F s at ai These coef cients are readily found with the aid of the rst two terms in the Taylor series for s about s0 ai 1 1 ms m M s my ai 1 ai s ai 160 MMm Sal2 Sai O ltls alllt 2a It is straightforward to show that Mai 12 and ai 0 and we nd that b1 0 and b2 12 Hence 13303 e31 Fm liens e Fs 2Reei 1 tcos at We can then conclude that ft tcosat a gt 0 provided that F s satis es the desired boundedness condition As for that 00ndition when z is apoint on CR lzlyRei9 s y R R 7 and IzllyRe 9I2Iy RI R ygt a and this means that Iz2 azl s lzl2 a 5 R y2 a2 and lz2 a2l 2 llzl2 a2l 2 R y2 a2 gt 0 Hence R 2 2 7a2gtOasR gtoo mm 6 We are given sinhxs F O lt lt 1 s s2 coshs x which has isolated singularities at the points so 0 Sn E39111 Ei and 5quot Q l l i n12quot 2 2 This function has the property FE F E and so m lies ms 2Res ems Res emu 30 quot1 35 15quot To find the residue at s0 O we write sinhxs xsxs 31 xx3s26 7c 2 2 2 3 O lt Isl lt s coshs s 1s 2n ss l2m 161 Division of series then reveals that so is a simple pole of Fs with residue x and according to expression 3 Sec 82 Res e Fs Res Fs x As for the residues of Fs at the singular points squot n 12 we write 175 2 Fs q s where ps s1nhxs and qs s coshs We note that 2n 1mc psisin 0 and qs0 furthermore since q s 2s cosh s s2 sinh s we find that 2 2 2 2 q s 2quot 41 isinan 1 i 2quot 41 sinmz 72 2n 12 22 2n 127rz n 4 4 1l 0 7r 71 sm nncos cosmrsm 2 2 In view of Theorem 2 in Sec 69 then 3 is a simple pole of F s and R F PS l4 1quot 2n 17x is S q s 72 2n 12 8 2 39 Expression 4 Sec 82 now gives us 1328 Fltsgt1 fequot 1 REE 2 2quot21 xpli 2quot quot 8 quot T 1 2 sin 27 lnx cos 2n 17tt 7 2n l 2 2 162 Summing all of the above residues we arrive at the nal result 8 quot 1quot 2n 1mc 2n lm f t x 72 quot1 2quot D2 sm 2 cos 2 7 The function F s scoshs 2 2 does not lie along the negative real axis has 2n 12 72 4 where it is agreed that the branch cut of s 2 isolated singularities at so 0 and when coshs O or at the points 5 n 12 The point so is a simple pole of Fs as is seen by writing 1 1 1 scoshs 2 quot s1sm22s1244 ss2 2s3 24 and dividing this last denominator into 1 In fact the residue is found to be 1 and expression 3 Sec 82 tells us that Res equotFs Res Fs 1 As for the other singularities we write F s 92 where ps 1 and qs scoshs 613 12 Now psl 0 and qsn0 also since q s s 2 sinhs 2 coshs 2 it is straightforward to show that 2n17r 2n l qsn 4 smn7r 2 4 1 0 163 So each point sl is a simple pole of Fs and ResFs p 2 3 q sn 7 2n 1 Consequently according to expression 3 Sec 82 s x 4 1quot 2n 12 zt R 39 F 5 Res F 3 e s e 3 e m s 7 2n 1 xp 4 71 12 Finally then f t lies e Fs E Res e Fs 30 quot1 or 1 12quot a ft 1 n2nlexp 4 Here we are given the function coth7rs 2 cosh7rs 2 F r S 52 1 32 1sinh7ts 2 which has the prOperty m F E We consider first the singularities at s ii Upon writing coshn39s2 992 Fs s i Where 3 sisinh7rs2 we nd that since i 0 the point i is a removable singularity of F 3 see Exercise 3b Sec 65 and the same is true of the point i At each of these points it follows that the residue of equotFs is 0 The other singularities occur when 71532 mti n 0tliZ or at the points s 271139 n OIliZ To nd the residues we write Fs 5 where ps 005 and qs s2 1sinh 7 164 and note that p2m39 coshmri cosmr 1quot a 0 and q2ni 0 Furthermore since 2 1 shm 2 q 539 S 2 CO 2 S S 2 we have 7174n2 q 2ni 4n2 1coshnm39 4n2 1 cosmr 2 39n I 1quot 0 Thus Resp fg39ll 21 21 s2m q 2nl 7 4n 1 n 0tli2 Expressions 3 and 4 in Sec 82 now tell us that IRese Fs Res 113 3 0 39 r0 7t and 2 I 4 cos 2m R quotF Res e F 2Re 2quot 12 sgnsi e 2ni e 7 4n2 l n 4 2 1 n The desired function of t is then 2 4 quot cos 2nt t f 7t 7 quot1 4 22 1 9 The function sinhxs 2 F s 0 lt x lt1 s2 sinhs 2 where it is agreed that the branch cut of as 2 does not lie along the negative real axis has 12 2 isolated singularities at s0 and when sinhs 0 or at the points s n27r n 12 The point 3 0 is a pole of order 2 of F s as is seen by first writing sinhxs 2 xs 2 xsmr I 3xs 25 5 xx3sl6x5s2 120 slsinho s2s 2s1233s 255 szsal6s4120 10 165 and dividing the series in the denominator into the series in the numerator The result is sinhxsm 1 1 3 x s2 sinhs 2 32 6 x x O ltsllt 112 In view of expression 1 Sec 82 then It 1 3 l 2 1330s e Fs g0 x xt Exx 1 xt As for the singularities s n17172 n 12 we write Fs l s 4 Observe that p n27t2 0 and q n27r2 0 Also since where ps sinhxs 2 and qs s2 sinhs 2 q s 2s sinhs 2 in cosms it is easy to see that q n27t2 0 So the points s 2272 n 12 are simple poles of Fs and M5 25inhxsquot2 2 quot1 Res Fs 2 zz qs n2xz S S112 cosh 312 2 a n3 smnmc n 1 Thus in view of expression 3 Sec 82 u 2 n1 2 z Essie Fs 11Te39quot quot 39sinmux n 12 Finally since ft 1330s e Fs 235 2equotFs we arrive at the expression 1 2 2 quot391quot1 4112 ft EXCX lxt32 l3 e 811171712Xquot 111 The function 1 1 PS s2 ssinhs has isolated singularities at the points s00 and snmri n nm nl2 166 Now ssinhssss3ms2 s4m 0 ltlslt and division of this series into 1 reveals that 1 1 1 1 Fm 2quot Far 64 6 0 ltlsllt 7 This shows that F s has a removable singularity at so Evidently then e Fs must also have a removable singularity there and so Res e Fs 0 To nd the residue of F s at s39 niri n 12 we write Fs 92 where ps sinhs s and qs s2 sinhs qs and observe that pom quotm at 039 90quot 0 and 61 nm39 n22391721quot1 e 0 Consequently F s has a simple pole at squot and pnn39i nn i 1 R F 12 3 S q mzi n2n21quot1 m n Since W3 F the points 5 are also simple poles of F s and we may write Res equotFs Res e Fs 2 Re nie quot marlquot icosnm 8111117150 33 i3 m n 2 sin mrt 1141 mt Hence the desired result is ft 135 equotFs fig equotFs 13 equotFs n1 on n1 m zz2H sinmrt 167 11 We consider here the function sinhxs F S ss2 w2coshs O lt x lt l 2 1 where a gt O and a 5 0 LILE n 12 The singularities of F s are at s O s icoi and s iconi n 12 Because the rst term in the Maclaurin series for sinhxs is xs it is easy to see that s 0 is a removable singularity of equotFs and that Res e Fs 0 To nd the residue of F s at s wi we write M sitlh06L Fs s au39 Where Ms ss coicoshs from which it follows that s coi is simple pole and sinhxcoi isinazx wi2wi coshai 2co2 cos a 39 1333 F S Mali Since 757 Fs then isinazx sinazx sinazxsinwt Res equotFs Res e 39Fs 2 Re 2 m 2 at 2 mi man 2a cos 0 26 cos a 0 cos a As for the residues at s Uni n 12 we put F s in the form F s 3 where ps sinhxs and qs s3 wzscoshs q s Now pai sinhani i sin aux at O and qwni 0 Also since q s s3 cozssinhs 3s2 12coshs we nd that q wi wji w wnisinhcoi on a2 cofsin aquot at 0 Hence we have a simple pole at s uni with residue Rams p at 1311160an I mm q cont aw ausm aquot 168 Consequently Res eFs estFs 2 Re lsm wnx eiwlt Sm wnx Sin out mmquot u 2 2 2 2 39 39 conaJ asmw calm cos1ncon But sin aquot sinmr 1 and this means that n1 39 39 Res eFS Res enFs 2L sm ngsmzwnt39 30quot 391 on 60 on Finally f t Res equotFs Res e Fs Res equot F s iRes e Fs Res equotFs 30 30 rzax quot1 10 ax That is 39 on n1 sm wxsxn cot 22 1 I sm conx sm cont 02 cos co m a a2 a ll f0 Midterm 2 sample problems DO NOT USE THIS SET OF SAMPLE PROBLEMS AS A STUDY GUIDE I MAY ASK QUESTIONS ABOUT OTHER TOPICS AS WELL DO THESE PROBLEMS AFTER YOU HAVE FINISHED STUDYING FOR THE EXAM Problem 1 Evaluate the following integrals 22 7 22 dz 0 along the positively oriented circle C 2 5 Logz d2 0 along the polygonal line C EFG Where E017 Fl I7 C12 a Problem 2 Evaluate the integrals 2 a 0 dz Where C is the positively oriented circle of radius 2 With center 0 b C dz Where C is the positively oriented circle of radius 2 With center 2 Problem 3 a Obtain the Maclaurin expansion Le7 the Taylor expansion about 20 0 of 2coshz2 b Let Find flt201 Hint nd the Taylor series about 20 1 of 2715373 and differentiate the series Problem 4 a Expand the function I 2 271w 73gt into a Laurent series in the domain 0 lt 2 7 1 lt 2 b Use division to obtain the rst 3 nonzero terms of the Laurent expansion of the function on the domain 0 lt lt l 2205271 Problem 5 a Find the maximum value of I cos 2 on the region R 2 E C 0 S Rez g 2W0 g Imz S 2 b Let C be the straight line segment that connects 02 and 20 Show that e 7dz S e C 2 Exam 1Sample DO NOT USE THIS SAMPLE EXAM AS A STUDY GUIDE ONLY TRY THIS EXAM AFTER YOU HAVE STUDIED ALL MATERIAL ON THE ACTUAL MIDTERM THERE CAN BE QUESTIONS ABOUT OTHER TOPICS AS WELL MATH 322 October 2010 Name No calculators or other electronic messaging devices7 as well as notes7 are permitted You may use the textbook Show all work for credit This is a 50 minutes exam with 5 problems Problem 1 Evaluate a Re a 1m 5 M 1 n 3 Problem 2 Sketch each of the following sets and determine whether the given set is open and0r closed a Rezlmzgt0 b 2lt lziilg3 l 3 c 51 lRez39lg27n39ltlmzlt77r Problem 3 Find the limits and prove the answer 22 i a zgrin 39 1 127 10 4 b lim 721 1 Z 0012372210 5 Problem 4 Find all the points z z iy if any7 where the function fz Mm y ivz y is differentiable Find f z at these points a HZ 902 7 M 0 Hz i i 7 N c Is any of these functions analytic If s07 nd the points where f is analytic Problem 5 a Using the de nitions of sinz and cos 2 and the de nition of tangent as tanz show that 17 52m tanw 2m 10 Using the formula in a7 show that tanz 7r tanz for any 2 E C

### BOOM! Enjoy Your Free Notes!

We've added these Notes to your profile, click here to view them now.

### You're already Subscribed!

Looks like you've already subscribed to StudySoup, you won't need to purchase another subscription to get this material. To access this material simply click 'View Full Document'

## Why people love StudySoup

#### "There's no way I would have passed my Organic Chemistry class this semester without the notes and study guides I got from StudySoup."

#### "I bought an awesome study guide, which helped me get an A in my Math 34B class this quarter!"

#### "There's no way I would have passed my Organic Chemistry class this semester without the notes and study guides I got from StudySoup."

#### "It's a great way for students to improve their educational experience and it seemed like a product that everybody wants, so all the people participating are winning."

### Refund Policy

#### STUDYSOUP CANCELLATION POLICY

All subscriptions to StudySoup are paid in full at the time of subscribing. To change your credit card information or to cancel your subscription, go to "Edit Settings". All credit card information will be available there. If you should decide to cancel your subscription, it will continue to be valid until the next payment period, as all payments for the current period were made in advance. For special circumstances, please email support@studysoup.com

#### STUDYSOUP REFUND POLICY

StudySoup has more than 1 million course-specific study resources to help students study smarter. If you’re having trouble finding what you’re looking for, our customer support team can help you find what you need! Feel free to contact them here: support@studysoup.com

Recurring Subscriptions: If you have canceled your recurring subscription on the day of renewal and have not downloaded any documents, you may request a refund by submitting an email to support@studysoup.com

Satisfaction Guarantee: If you’re not satisfied with your subscription, you can contact us for further help. Contact must be made within 3 business days of your subscription purchase and your refund request will be subject for review.

Please Note: Refunds can never be provided more than 30 days after the initial purchase date regardless of your activity on the site.