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## Complex Analysis

by: Gunner Price III

16

0

7

# Complex Analysis MATH 412

Gunner Price III
Cal State Fullerton
GPA 3.92

Scott Annin

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COURSE
PROF.
Scott Annin
TYPE
Class Notes
PAGES
7
WORDS
KARMA
25 ?

## Popular in Mathematics (M)

This 7 page Class Notes was uploaded by Gunner Price III on Wednesday September 30, 2015. The Class Notes belongs to MATH 412 at California State University - Fullerton taught by Scott Annin in Fall. Since its upload, it has received 16 views. For similar materials see /class/217023/math-412-california-state-university-fullerton in Mathematics (M) at California State University - Fullerton.

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Date Created: 09/30/15
Math 412 Group Work 1 Spring 2009 SOLUTIONS Problem 1 Evaluate each of the following a7na7m a 71 77 i SOLUTION We have 47i1732 7 1713i 7 1713i7172z 7 72711i 727 11 7 72 7l2i 7l2i 712i7172i 5 5 5 D l bI gt mltziwgt SOLUTION We have 1 zzy I 7 y 7 72 Iiiy z2y2 z2y2 124w sowehave I 1 7 y m z7iy 712y2 D Problem 2 In the complex plane7 sketch the points determined by a l2 l 7 22quot 2 SOLUTION This is a circle of radius 2 in the complex plane and centered at 717 2 the complex number 71 2239 Thus7 2 z yi is on the circle in question if and only if 1 12 y 7 22 4 D b Imz 7 2239 gt 6 SOLUTION The complex number 2 abi satis es Imz 7 2239 gt 6 if and only if b 7 2 gt 6 That is7 b gt 8 Hence7 the set in question consists of all points in the complex plane above the horizontal line y 8 D Problem 3 Show that a 21 22l2l21lil22l SOLUTION By the Triangle Inequality7 we have 21 21 7 22 22 S 21 7 22 22 and a simple rearrangement of this inequality yields the desired result D SOLUTION Write 21 a1 121239 and 22 a2 1222397 Where a1 1217a27 122 E R Then 271 a1 121239 a2 7 122239 a1a2 121122 a2121 7 a1122 22 a 12 a 12 39 a 12 Hence7 1 7 MW 111112 a2bi 1112 7 a3 123 a b 39 2392 On the other hand7 E 7 a1 7 121239 a1 7 121239 a2 122239 a1a2 121122 a1122 7 a2121 2221271222 a b a b a b A quick inspection shows the desired equality Math 412 Group Work 11 Spring 2009 SOLUTIONS Problem 1 In class we derived the formula arcsinz 7239 log 2392 17 2512 Using the same procedure yes you can peek at your notes derive the formula arccos z 7239 log 2 22 7112 SOLUTION To derive the formula for arccosz we must solve z cos w for w To do this we begin by writing 2 cosw em 6 Hence I I 6 7 22 671w 0 Multiplying through by 6quot we obtain em 7 226quot 1 0 Viewing this as a quadratic equation in the unknown77 quantity 6quot we can apply the quadratic formula to obtain 7 22 422 i 4W2 7 7 2 7 Taking the logarithm of both sides this becomes eiw z 22 7112 z w log 2 22 7112 w 7z39log z 22 7112 which is precisely the desired formula for arccos z D Problem 2 Differentiating both sides of the equation 2 cosarccos z with respect to z derive the formula 1 d 7arccos z 7m dz SOLUTION Di erentiating both sides of the equation 2 cosarccos z with re spect to 2 we have 1 7 sinarccos z arccos z 7 so that 1 i 7 COOS Z i sinarccos Letting w arccosz7 we have cos w 2 so that sinZ w 17227 or sinw 172212 That is7 sinarccos z 1 7 22 127 from which we arrive at 1 arccos z 7m Problem 3 Find all values of arccos 3239 SOLUTION From Problem 17 we have arccos 3239 7239 log 3239 71012 There are two values for 710127 namely x10239 and ixloi and both must be con sidered For x102397 we must compute 410g 3239 m2 410g 3 my 7239 1113 xE z39 27mgt 7r 2n 7 1n3 m For 7v10239 we must compute 410g 3239 7 410g 4 7 3n 7 ln1i 7 3 239 27m 7r 32n 4111W73 Math 412 March 13 2009 Lecture Spring 2009 Dear Math 412 Students I decided that my lecture on Friday March 13 was not my nest hour of explaining mathematics partially because of the arge amount of similarsounding but not quite the same factsl Since I donlt want to have to wait a whole weekend to make things better l7m going to try and summarize and clarify some things from Section 41 with this handout Here are the key relationships to know 1 If converges then converges Converse false 2 71 2 The converse of the above does hold in the special case when the convergence is happening at 0 That is A 0 if and only if A 0 oo 3 If 22 converges then A 0 Converse false 2 n1 This says that if the series converges then the corresponding sequence converges too Not only that the corresponding sequence must converge to 0 Therefore the sequence also converges to 0 by Point 2 4 oo oo If E converges then E 2 converges mil 7 711 71 TL oo oo n l Converse false Use 2 E 1 converges but E 1 does not n n7 This says EXACTLY NO MORE AND NO LESS THAN THE FACT that every absolutely oo convergent series 22 is convergent n1 Remark Notice Points 1 and 4 carefully Neither converse is true and the statements go in opposite directions Donlt let this confuse you 2 In our example at the end of class 2 r in we saw that l for all n so that the sequence converged However the sequence did not converge the terms of the sequence ounce around on the circle of radius 1 in the complex plane Thus we have another Converse false77 example for Point 1 above In particular the sequence certainly did not converge to 0 so by Point 3 above the series 00 oo 22 did not converge and so by Point 4 Z did not converge eitherl Bottom line If n1 n1 the terms do not go to zero forget about having any kind of a convergent series consisting of those terms Got it Good Now lets do another example 34z39 n 6 gt converges absolutely converges conditionally or diverges oo Decide Whether lt 7 Solution We have 2 554439 Now 262916 6 6 7 36 6 Thus for all n Thus the sequence converges to zero and therefore by Point 2 the sequence also converges to zero 34i 6 Now what about the series Since the converse of Point 3 fails we cannot use it to draw any conclusions Perhaps you remember from calculus that if z E R with lt 1 then 1 1z12m7l 171 00 In other words a geometric series that uses 1 such that lt l converges Since is a 711 00 geometric series and we veri ed that S g g lt 1 already we conclude that Z convergesl n 1 00 Hence 22 CONVERGES ABSOLUTELY Point 4 D 711 We will have more tests for convergence of a seriesiso don7t worryl You ll be getting more tools soonl Enjoy the weekend Scott Math 412 Group Work 11 Spring 2009 Problem 1 ln Class7 we derived the formula aresinz filog 2392 1 2512 Using the same procedure yes7 you can peek at your notes derive the formula arccosz filog z 22 11239 Problem 2 Di erentiating both sides of the equation 2 cosarccos z with respect to z derive the formula d 7 1 d2 arccosz i 1722 Problem 3 Find all values of arecos 3239

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