sfinal - MA 412 Complex Analysis Final Exam Summer II...

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MA 412 Complex Analysis Final Exam Summer II Session, August 9, 2001. 1. Find all the values of ( 8 i ) 1 / 3 . Sketch the solutions. 1pt . Answer: We start by writing 8 i in polar form and then we’ll compute the cubic root: ( 8 i ) 1 / 3 =( 8 e iπ/ 2 ) 1 / 3 =8 1 / 3 exp ( i ( π 6 + 2 πk 3 ) ) , Hence z 0 =2exp( iπ/ 6), z 1 iπ/ 2) = 2 i , z 2 i 7 π/ 6). 2. Suppose v is harmonic conjugated to u ,and u is harmonic conjugated to v . Show that u and v must be constant functions. . Answer By de±nition, v is harmonic conjugated to u if Δ u v =0and u x = v y ,u y = v x hold. On the other hand, as u is harmonic conjugated to v , we also have v x = u y and v y = u x . Then, u y = v x = v x implies v x u y =0 and u x = v y = v y implies v y u x . Hence, u ( x, y )and v ( x, y ) are constant functions. 3. Show that f ( z )= | z | 2 is di²erentiable at the point z 0 = 0 but not at any other point. . 1
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z 2 z 0 z 1 2 = 2i Figure 1: Roots of ( 8 i ) 1 / 3 . Answer: We could show that f is not di±erentiable at any z 0 C 0by computing di±erent limits for di±erent trajectories. But we’ll use the necessary conditions of di±erentiability ( i.e. the Cauchy-Riemann equations) to do so. First, we write f ( z )= u ( x, y )+ iv ( x, y ), so u ( x, y x 2 + y 2 and v ( x, y )=0. Clearly, the ²rst-order partials of u and v are continuous everywhere. But u x =2 x equals v y =0 , and u y y equals v x if and only if x = y = 0. Then f
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This note was uploaded on 06/17/2011 for the course MATH 132 taught by Professor Grossman during the Spring '08 term at UCLA.

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sfinal - MA 412 Complex Analysis Final Exam Summer II...

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