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# Complex Variables MAT 3223

UTSA

GPA 3.84

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This 35 page Class Notes was uploaded by Miss Cloyd Cronin on Thursday October 29, 2015. The Class Notes belongs to MAT 3223 at University of Texas at San Antonio taught by Dmitry Gokhman in Fall. Since its upload, it has received 26 views. For similar materials see /class/231335/mat-3223-university-of-texas-at-san-antonio in Mathematics (M) at University of Texas at San Antonio.

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Date Created: 10/29/15

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0 FrMaL 139 pas5L 1h 7m3939c if W0 Y 4 All Janis 7M 4 Ts W39tcn m 5 quotelm Lu H53 39 8k in 1 s a 5 QM Hal 4 391 M 41th o 1 A96 duty 1 QMY MM Uinlin 39 iq Hin954139t J S 11 is 2T Y 39 h was Btrhmrn39 P I m lt R lhe w b 4m ZahLL up a1 p 3quot o Wt h 39HM ak M mman 3 29 v 9 Wot Hui a 3 Hate anm m gmm M M39v 4ko N 39e39 M h AH Mn Guns p r lamay 4 Jab hId 39hm II 4 zisclm ga w J mm MOI I pineal 39Fa39 h l quotl naer M r LG lIl Liv Ra at 24p hotr aa 931 WEN 2 mru a IiFYquot Kanlymph Chapter Three Elementary Functions 31 Introduction Complex functions are of course quite easy to come byithey are simply ordered pairs of realvalued functions of two variables We have however already seen enough to realize that it is those complex functions that are differentiable that are the most interesting It was important in our invention of the complex numbers that these new numbers in some sense included the old real numbersiin other words we extended the reals We shall find it most useful and pro table to do a similar thing with many of the familiar real functions That is we seek complex functions such that when restricted to the reals are familiar real functions As we have seen the extension of polynomials and rational functions to complex functions is easy we simply change x s to z s Thus for instance the function f defined by 22Zl z7 21 has a derivative at each point of its domain and for z x 0139 becomes a familiar real rational function x2xl fooi xl 39 What happens with the trigonometric functions exponentials logarithms etc is not so obvious Let us begin 32 The exponential function Let the socalled exponential function exp be defined by expz ecosy z39siny where as usual 2 x z39y From the CauchyRiemann equations we see at once that this function has a derivative every whereiit is an entire function Moreover iex z expz Note next that if z x z39y and w u z39v then 31 expz w cosy v 1 Smy v exeweosyeosv Smysmv Ismycosv cosyslnv cosyx slnycosv tslnv KW EXPW the real enponentlal ax to the complex nurnbers Example Recall from elementary elreult analysls that the relatlon between the voltage olrop Vanol the z VL L Frntnrlutrth a lnoluetanee and for a eapaeltor c 1 where Us the eapaeltanee The vanable tls of eourse urne Note that l Vls slnusoldal wlth a en Suppose then that V Aslnmt 1 We ean wnte tlus as where 3 rs complex We know the eunent1wlll have thls CE W The relataons between the voltage anolthe eurrent are hnear anol voltages and eurrents and use the fact that w eosmn Ismail We thus assurne arnore or less fletlonal complex voltage V the lrnaglnarv part ofwhlch ls the aetual voltage and then the aetual eurrentwlll be the lrnaglnarv part of the resultlng complex eurrent relatlon ls V Ll where L ls the slrnplv rnulupheatlon by In Ae tholaw If AW the above relataons between urrent and voltage beeorne V mJLIfor anlnoluetor andme 1 or V for a eapa tor T tllu tr t RLC elreult wlth avoltage souree V aslnmt We let E Em E L gt en the faetthat the voltage olrop arounol a eloseol elreult rnustbe zero one of Krreholl s Th eel ebratedlaws looks llke 1011 R gemquot or sz iwLb bC Rb a 1390 Thus b RiwL In polar form b a elkJ R2 coL if a where coL LC tanrp Hence I Imbequot 1m emmw R2 coL ll C sinat p This result is wellknown to all but I hope you are convinced that this algebraic approach afforded us by the use of complex numbers is far easier than solving the differential equation You should note that this method yields the steady state solutionithe transient solution is not necessarily sinusoidal Exercises 1 Show that eXpz 2m eXpz expltzgt 2 Show that BMW 7 eXpz w 3 Show that leXpzl ex and argeXpz y 21a for any argeXpz and some 33 rhtegerk 4 Fmd all z sueh that expz 71 or explarh why there are none 5 Fmd all z sueh that expz 1 Lor explam why there are none 51 r 7 End the mdmated mesh eurrehts m the network where we are usmg Ex expz Frrst1et39s ve So supposez m 01 X Then w cosxxsxnx and osxexsmx ix fix 005x ix fix smx and everything is just ne Next observe that the sine and cosine functions are entireithey are simply linear combinations of the entire functions 2 and 2quot Moreover we see that i sinz cosz and i cosz sinz dz dz just as we would hope It may not have been clear to you back in elementary calculus what the socalled hyperbolic sine and cosine functions had to do with the ordinary sine and cosine functions Now perhaps it will be evident Recall that for real I f if f 4 s1nht and cosht Thus smm 1mmr Similarly cosz39t cosh I How nice Most of the identities you learned in the 3 01 grade for the real sine and cosine functions are also valid in the general compleX case Let s look at some sinzz coszz e z 2 e e402 i 4 62iz Zeizeiiz 672 eZiz Zeizeiiz 7212 ii 740m 1 35 It is also relative straightforward and easy to show that sinz i w sinzcosw i coszsinw and cosz i w coszcosw 4 sinzsinw Other familiar ones follow from these in the usual elementary school trigonometry fashion Let s find the real and imaginary parts of these functions sinz sinx z39y sinxcosz39y cosxsinz39y sinxcoshy z39cosx sinhy In the same way we get cosz cosxcoshy z39sinxsinhy Exercises 8 Show that for all z asinz 27 sinz bcosz 27 cosz csinz cosz 9 Show that lsinzl2 sinzx sinhzy and lcoszl2 coszx sinhzy 10 Find all 2 such that sinz 0 11 Find all 2 such that cosz 2 or eXplain why there are none 34 Logaritth and complex exponents In the case of real functions the logarithm function was simply the inverse of the exponential function Life is more complicated in the compleX caseias we have seen the compleX eXponential function is not invertible There are many solutions to the equation ez w Ifz 0 we de ne logz by logz lnlzl z39argz There are thus many logz s one for each argument of z The difference between any two of these is thus an integral multiple of 2m First for any value of logz we have 36 elogz elnlzlH39argz elnlzleiargz Z This is familiar But next there is a slight complication logez lne z39argez x y 2k7rz39 Z 2km where k is an integer We also have logzw lnlewl z39argzw lnlzl z39argz 1n lwl z39argw 2km logz logw 2km for some integer k There is de ned a function called the principal logarithm or principal branch of the logarithm function given by Logz lnlzl z39Arg z where Arg 2 is the principal argument of 2 Observe that for any logz it is true that logz Log 2 2km for some integer k which depends on 2 This new function is an extension of the real logarithm function Log x lnx z39Arg x lnx This function is analytic at a lot of places First note that it is not defined at z 0 and is not continuous anywhere on the negative real axis 2 x 0139 where x lt 0 So let s suppose 20 x0 z39yo where 20 is not zero or on the negative real aXis and see about a derivative of Log 2 Log isoog 20 lim Log 2 Log 20 lim z gtz0 eLogz eLogzo z gtz 0 Now if we let w Log 2 and wo Log 20 and notice that w gt wo as z gt 20 this becomes 37 Logz Logzo w wo 11m 11m z gtz0 Z 20 w gtw0 e e 0 1 L 8WD 20 Thus Log is differentiable at 20 and its derivative is We are now ready to give meaning to z where c is a complex number We do the obvious and de ne Zc eclogz There are many values of logz and so there can be many values of 25 As one might guess 25 is called the principal value of 25 Note that we are faced with two different de nitions of z in case 0 is an integer Let s see if we have anything to unleam Suppose c is simply an integer c n Then Zn enlogz enLogz2kni enLogzezkmu39 enLogz There is thusjust one value ofzquot and it is exactly what it should be e iogz lzlquotequotquotalgz It is easy to verify that in case 0 is a rational number 25 is also exactly what it should be Far more serious is the fact that we are faced with con icting de nitions of z in case 2 e In the above discussion we have assumed that ez stands for expz Now we have a de nition for ez that implies that ez can have many values For instance if someone runs at you in the night and hands you a note with 212 written on it how to you know whether this means expl 2 or the two values J and J Strictly speaking you do not know This ambiguity could be avoided of course by always using the notation expz for exeiy but almost everybody in the world uses ez with the understanding that this is expz or equivalently the principal value of ez This will be our practice Exercises 12 Is the collection of all values of logz39 12 the same as the collection of all values of logz Explain 13 Is the collection of all values of logz392 the same as the collection of all values of Zlogz39 Explain 38 Dxuote39ime a 4 1120053 A M 39Lin mduwh39w AM Hm JggpvI TEET 3911 quotHA er NH rluc Vnhalf phnc Mt c f D 16 JIMdim adv40 Jr q Ha z 0 Yuk 395 P ns H WJVV WZTaIQ39Hz39 Hzi I j Lb 1 05 lquot 13H39Zu 3 39 2 39gtI haoe qr n 2 0 Q W bh AJPL NnBZ Tt 2quot 111 39E h quotL 7 Iz kht a39 139 neth 21 22 2 1 13 3 2quot 2L LP 73ch C c39 QCh 1 13 4 xi zr 2 4z TWMR 4 39 61 quot C TU 1 2L c4l ii3i 4 quoth h 1MB Hume Dvrs S rcvhu E VH L 23 C HUl J39dr d 1 39S at 239 E sl7 T1 2 zc I t 11i 3 z 4 I39ln luw mlmll lthruI cul MIN E T We 9 N 110 l l P w 7TH ll WITN 9Tfk Ill H II N r ww N1 T lwir Nn 43 H II n n Mr N DINKh rt Q qlndq u r2 7 if C th weer 239 WK VP Celt ah s J 23 a 1 dIUu sm 13 31 r qr L were 11 mvu m E WI n WIth T I 9322 Hscoak M quot ht th WE n e l 0 l N Pn 0 Hr Mb 2395 L T p 2 EFL Sea Fanan I levw C m Ayn n3 1 L 93 2 sF w A Ow 3 find pup o r fan 13 Nt l when C K NTSQI TNc n 3an CarAble 4 Z 239 3 i393H3znwmm 24 V J l 9 c 3q s u Wit t 3L 39 l 12quot Qm 0 5 QH i 3 M 16 i Q2L 145 44 7 34 L a 11 FL QCMX DL J Q Ii 1 i L 9 L3H 14r 39 39 C 10 QK39DYEDz Mg 4an Fol 7w L r an MHJH Nun Id v r ATTHQHI is Visfwn l nm5no u nEL Slag n Hana a 1er I rst 3 quot4 a r For n 4 C Bffs 55 mnrd arm v UN Wanna wh T WVHOO g V if n lr ur EL Ir 1 1 nunqu n w 7 F o Y Vii I L 1 PT I F 56 P3 0 l lt P nusLFL fr 4 Frn w Mn 4 w i ikar pnh u l 4L9 R 3 f 117 a 1 Q Ha I 9quotan 9 059 Cu akakl VikkkSIb o D952 9 H4 I wn quotHawk teCD 0 Mn WM 392 in C4610 Snu 403 1 k rm 6 52m 3 be h ww w1csfdm K k iw LA a 4m Lu chm ce IR Vrck cgtl 39 cRM Maker mmcc m lt J 0ltClt meow 4 c 4 1 its Sinu cFade MM 2 m e b s39 941nm 4x20 r39h3 l quot quot a 54 gt l dbw A uh um I km 5 139 39 312009 Tiaras em 1 DxuoLcr Tnie M Dev 439 m w 6 a 7 WEMVquot Hrlwlu Jlan u a raft0 maPad 211 V mku v Q In madm 4 u Ma slF Q I Vb cer M39h 4 use arrb q 1 J h ctr c a 3 o x 1 filth 2 9 garb 1 9 Q c 3 I 04th 44 Q aw ol n5 SawL Lu 94 A V 14 D39 l 4 LL QtP 14c 4 LJ I It WSW h cu a c 0quot n3 OHM639 T Link 3 re 5 PH TS FE 73quot apkq3wesb CE1 Cvicvlrdsal n5 fthT24 qqug trtruly c1411 A v 039 n39q Lr A l5 L4 rsl39x t wdm 4As WYoS39Iln M q MBLIma MJJWVHAL39JMJ omurmh 39h wH39x vHylukr IvaiuAn39u LIEUI W 39 NDLWA h 392 NHM MAJlulu M Akbro sunr wdh m MEL Munckh m invukuc us mkx invmm MOIMWIS W ynlra39hl Su w M39J39l39 Hai39 M lH39vHL MI human 2 1 M5 Hus M wukm ba llamw Um Lam 4 untaaf 660 thus 1 2 r E tav rL Qi z dio r1 Z 39t 5br1 3 Chapter Fnur Integration 41 Intrndllctinn Ify D a c 15 srrhply afuheuoh on areal interval D m5 thenthe u integral Mod rs ofcourse srrhply an ordered pan of everyday 3394 grade calculus integrals u u u 012 Jxtdt 1Jytdt where yt r0 W Thus for example r 2 2 ii 1 Jum 1xt 11 34 c e D2C ane 39 to two dJmEnSlonS ofwhatwe dadm one dJmEnSlon back m Mrs Tumer s class In one d w w r u get from one numberto the other here we must also speerfy apath from a to A Let Cbe a n m Fr m V but th case a e A s We estn apath 0t Hr quotd W n tut V th m Next 1etP be apartitinn ofthe curve that ts P 251112 2 ts afthtte subset of c sueh that a zu b z audsuehthatz comes tmmedtatety afterzkx as we travel along Cfrom a to A Rtemauh sum assoetated wtth the partition Ptstustwhattt ts ththe test1 case 30quot E 72AZJ t whetez ts apotht on the are between zt audz and A2 z ezrt Note thatfor a gtveh partition P thete ate many SltAdepeh hg on how the pothts are chosen If thete ts ahumbet L sothatgtveh any s gt 0 thete ts apathtton PE ofC sueh that mew e everP 3 PE thehts satdto be thtegtabte on c and the humbetL ts estnedthe integral nff on c Thts humbet L ts usually wutteh Matt c Jszdz sJzdz c c at any complex constant c J39z gzdz J39 zdz J39gzdz C C C 42 Evaluating integrals Now how on Earth do we ever nd such an integral Let y 11 gt C be a complex description of the curve C We partition C by partitioning the interval 11 in the usual way a to lt t1 lt t2 ltlt tn Then a yayt1yt2 y b is partition ofC Recall we assume that y t 0 for a complex description of a curve C A corresponding Riemann sum looks like SltPgt Z mpxm ml j1 We have chosen the points z 90 where IN S t S If Next multiply each term in the sum by l in disguise SltPgt mexWXo r171 11 I hope it is now reasonably convincing that in the limit we have 3 I zgtd2 meodn C a We are of course assuming that the derivative 9 exists Example We shall nd the integral of fz x2 y z39xy from a 0 to b l 139 along three different paths or contours as some call them First let C1 be the part of the parabola y x2 connecting the two points A complex description ofC1 is y1t t 112 0 5 t5 1 43 Now y 1t 1 2n andf y1t t2 t2 m2 2t2 113 Hence 1 New In mowed c1 0 l J392t2 mm 2tz39dt 0 1 Jth 2t4 5t3idt 0 u 1z U1 blm Next let s integrate along the straight line segment C 2 joining 0 and l 139 Here we have y2t t it 0 S t S 1 Thus y 2t l z39 and our integral looks like 44 1 New I mow zmdr C2 0 t2 0 z39t2l z39dt t z39t 2t2dt o o N o1 Finally let s integrate along C 3 the path consisting of the line segment from 0 to 1 together with the segment from 1 to l 139 We shall do this in two parts C31 the line from 0 to l and C32 the line from 1 to l 139 Then we have I fzdz J39 fzdz J39 fzdz C3 C31 C32 For C31 we have yt t 0 5 t5 1 Hence 1 7 2 7 i Jf2dzit dt7 3 C31 0 For C32 we have yt l it 0 5 t5 1 Hence 1 i l I z zi lltztzdt7 2 2 c32 0 45 Thus J39 zdz J39 fzdz J39 fzdz C3 C31 C32 Suppose there is a numberM so that l z S M for all 26C Then J39 zdz C 3 I mfm 3 SIWKOWKWW a 3 5 MIly tldt ML a l3 where L Ily tldt is the length of C Exercises 1 Evaluate the integral lid where C is the parabola y x2 from 0 to l z39 c 2 Evaluate I dz where C is the circle of radius 2 centered at 0 oriented c counterclockwise 4 Evaluate I z z where C is the curve y x3 from 1 z39 to l z39 and c 1 for ylt0 z 4y for y20 5 Let C be the part of the circle yt e in the rst quadrant from a l to b 139 Find as small an upper bound as you can forlJCz2 74 5dz 46 6 Evaluate I z z where fz z 27 and C is the path from z 0 to z l 2139 c consisting of the line segment from 0 to 1 together with the segment from 1 to l 2139 43 Antiderivatives Suppose D is a subset of the reals and y D gt C is differentiable at t Suppose further that g is differentiable at yt Then let s see about the derivative of the composition gyt It is in fact exactly what one would guess First g7t u961yt ivxtyt where gz ux y z39vx y and yt xt z39yt Then i Lug Elli ii ii dzgm ax d By d ax d By d 39 The places at which the functions on the righthand side of the equation are evaluated are obvious Now apply the CauchyRiemann equations gyt iii ii ii 6x dt 6x dt 6x dt 6x dt 6u amp i Q 6x adez dz g rtr t The nicest result in the world Now back to integrals LetF D gt C and suppose F z fz in D Suppose moreover that a and b are in D and that C C D is a contour from a to b Then 3 I zgtdz Iyrgtgty rgtdr C a where y 11 gt C describes C From our introductory discussion we know that F7t F 7t9 t 7t7 t Hence 47 3 I zgtd2 momma C a L3 F7tdt FWD F7 11 Fb Fa This is very pleasing Note that integral depends only on the points a and b and not at all on the path C We say the integral is path independent Observe that this is equivalent to saying that the integral of f around any closed path is 0 We have thus shown that if in D the integrand f is the derivative of a function F then any integral I z z for C C D is path c independent Example Let C be the curve y XL from the pointz l z39 to the pointz 3 Let s nd Izzdz C This is easyiwe know that F 2 22 where F 2 23 Thus 22512 z393 3 Now instead of assuming f has an antiderivative let us suppose that the integral off between any two points in the domain is independent of path and that f is continuous Assume also that every point in the domain D is an interior point of D and thatD is connected We shall see that in this case f has an antiderivative To do so let 20 be any point in D and de ne the function F by W Izwz Cz where C is any path in D from 20 to 2 Here is important that the integral is path independent otherwise F 2 would not be wellde ned Note also we need the assumption thatD is connected in order to be sure there always is at least one such path 48 Now forthe computation othe denvauve ofF Fz A2 7 72 I ew In Where LA 15 the hue segment from z to z A2 ZAZ Zo Next observe that 1 d5 A2 ThusJKz Mme andwe have z z Wem AZ Mmemy Now then s I lwwmaxmswrz 56L 1 Jars ezws s maxwtsr zn MA Weknowxs continuous atz and so hm mam542 56L 0 Henee M 2 W 712 21 t J ms erznds 1 0 In other words F z z and so just as promised f has an antiderivative Let s summarize what we have shown in this section Suppose f D gt C is continuous where D is connected and every point of D is an interior point Then f has an antiderivative if and only if the integral between any two points of D is path independent Exercises 7 Suppose C is any curve from 0 to 7 2139 Evaluate the integral tangy C 8 aLetFz logz 7r lt argz lt 7r Show that the derivative F z bLet Gz logz lt argz lt 74 Show that the derivative G z cLet C1 be a curve in the righthalf plane D1 z Rez 2 0 from 139 to 139 that does not pass through the origin Find the integral dLet C 2 be a curve in the lefthalf plane D2 z Rez S 0 from 139 to 139 that does not pass through the origin Find the integral 9 Let C be the circle of radius 1 centered at 0 with the clockwise orientation Find 1 J 7dZ c 10 aLet Hz 22 7 lt argz lt 7 Find the derivative H z bLetKz 25 lt argz lt 74 Find the derivative K z cLet C be any path from 1 to 1 that lies completely in the upper halfplane and does not pass through the origin Upper halfplane z Imz 2 0 Find 410 75 J f 7 h Aeraggrth 6 1 r h53 VII 1 rrlao r 9 EL PM 3 Q r 1 n i PL in 5 v m Ewhm m F 35 35 3 fr Saav 6 fl m gerbils Q WI if v n P i hrrmbn 1 i 3 J 6932 4 9 M 0L 5 abrimbx H 9 into go L a 15 n54 eclmfu F We Sima s y at J Y 5 9 Fix 4 95 e 5 4ng Dth m rift X 05 5 V 925 3 Sr in fwb rmdd f Amt v MGMV aw 2 Aman 422 N90quot J0 r y newwa r 4 vb hrwaio n rwaq Fr m m3quot Sliy Mrs 33149551 Id 0 Au in h 5t3 x 391 f r Ti H T 133 925 m me PL 3 3 9 53 66 1m b L We 23 v T5 r 31 N an xi u 7 33 n pars HH 3 C lt n 31ch u Mn UJlt71wvu Salvo snafu swolub SVHFBD erv Mc M 3a 7ab ltuuvrdrwm IJI 0on 69 i 51x r rrnfri 59 04 n 3 7 a 0549 wr r 3 1 a 23 L H i v h i r 9 34 9 w 2 I k 17933 n agn lob its SM 3395 nIF Tray Maw 9E Ag 540 tsz A 5 E 5 M x f 1 t baitquot xfrgxge u kw WW 5 lt gamma la 399 2 1 1 Eisem39texnquot 5 4 ie 439 cu r e1 1 01 l 11 5 H 7 ll glhz I rut7 u II 1 H In sill Lush A icm ll 9 in 6 i Kiiv quot H39ii P w 5 5 a c I 3 1L ax 997 e e ax fw c9m awnquot 5 f L LI 7 39739 7 339 39 00 e t 5 c

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