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Unformatted text preview: ACM 95a/100a Midterm Solutions Prepared by: Zhiyi Li November 10, 2007 Total: 122 points Include grading section: 2 points Problem 1 a) (6 pts) Give all possible values of f ( z ) = z in the form x + iy . b) (6 pts) Sketch the region of the complex plane corresponding to  z 2  <  z  . c) (6 pts) Find all the roots of cos z = e i/ 4 . d) (12 pts) Find a mapping w = f ( z ) that maps the strip 0 x < 1 with z = x + iy to the whole w plane (except for the origin) exactly once. Hint: use an appropriate periodic function. Solution 1 (a) (6 points) We can write f ( z ) = z = e log z = e Log  z  e i arg( z ) =  z  e i (Arg( z )+2 k ) =  z  cos ( Arg( z ) + 2 2 k ) + i  z  sin ( Arg( z ) + 2 2 k ) , k = 0 , 1 , 2 , . . . Note that z has an infinite number of branchesas we expect since is irrational. (b) (6 points) Let z 2 = r 2 e i 2 and z = re i . Then  z 2  <  z  = r 2 < r This implies < r < 1 or <  z  < 1 . The region is pictured in Figure 1. (c) (6 points) Using the definition cos z = 1 2 ( e iz + e iz ) , e iz + e iz = 2 e i 4 = e 2 iz 2 e i 4 e iz + 1 = 0 = e iz = e i 4 + ( e i 2 1 ) 1 / 2 by the quadratic formula. Take logarithms of both sides and simplify. z = 1 i log h e i 4 + ( e i 2 1 ) 1 / 2 i = 1 i log e i 4 + 2 e i 3 4 1 / 2 = 1 i log h e i 4 + 2 1 / 4 e i 3 8 i , 1 i log h e i 4 + 2 1 / 4 e i 5 8 i 1  z  < 1 1 1 i i Figure 1: The region for problem 1(b). Expanding the log, n 2 k + Arg e i 4 + 2 1 / 4 e i 3 8 i Log  e i 4 + 2 1 / 4 e i 3 8  , 2 k + Arg e i 4 + 2 1 / 4 e i 5 8 i Log  e i 4 + 2 1 / 4 e i 5 8  o , k = 0 , 1 , 2 , . . . (d) (12 points) First, we show that = e g maps infinite horizontal gstrips of width 2 to the whole plane (except for the origin). The strip R = { g : 0 < I ( g ) 2 , < R ( g ) < } (Figure 2) maps onetoone to the whole  plane (except for the origin). Let g = u + iv , also let ( , ) be the polar coordinate of = exp( g ). Consider four points a = x + 2 i , b = x + 2 i , c = x , d = x on the boundary of the strip R . The points a, b, c, d map to A, B, C, D . For point g = u + iv on rectangle abcd , since < u < and 0 < v 2 , we have exp( x ) = exp( u ) exp( x ), 0 < = v 2 , which shows that the image of the rectangle is contained in the annulus exp( x ) exp( x ). In the opposite direction, for a point on the annulus, its polar coordinate ( , ) must satisfy exp( x ) exp( x ), 0 < 2 . It is the image of g = log + i R under the map = e g ....
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 Fall '06
 NilesA.Pierce

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