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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 π 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 iπ/ 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 π = e π log z = e π Log  z  e iπ arg( z ) =  z  π e iπ (Arg( z )+2 πk ) =  z  π cos ( π Arg( z ) + 2 π 2 k ) + i  z  π sin ( π Arg( z ) + 2 π 2 k ) , k = 0 , ± 1 , ± 2 , . . . Note that z π has an infinite number of branches–as we expect since π is irrational. (b) (6 points) Let z 2 = r 2 e i 2 θ and ¯ z = re iθ . 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 π 4 = ⇒ e 2 iz 2 e i π 4 e iz + 1 = 0 = ⇒ e iz = e i π 4 + ( e i π 2 1 ) 1 / 2 by the quadratic formula. Take logarithms of both sides and simplify. z = 1 i log h e i π 4 + ( e i π 2 1 ) 1 / 2 i = 1 i log e i π 4 + √ 2 e i 3 π 4 1 / 2 = 1 i log h e i π 4 + 2 1 / 4 e i 3 π 8 i , 1 i log h e i π 4 + 2 1 / 4 e i 5 π 8 i 1  z  < 1 1 1 i i Figure 1: The region for problem 1(b). Expanding the log, n 2 πk + Arg e i π 4 + 2 1 / 4 e i 3 π 8 i Log  e i π 4 + 2 1 / 4 e i 3 π 8  , 2 πk + Arg e i π 4 + 2 1 / 4 e i 5 π 8 i Log  e i π 4 + 2 1 / 4 e i 5 π 8  o , k = 0 , ± 1 , ± 2 , . . . (d) (12 points) First, we show that ω = e g maps infinite horizontal gstrips of width 2 π to the whole ωplane (except for the origin). The strip R = { g : 0 < I ( g ) ≤ 2 π,∞ < R ( g ) < ∞} (Figure 2) maps onetoone to the whole ω plane (except for the origin). Let g = u + iv , also let ( ρ, φ ) be the polar coordinate of ω = exp( g ). Consider four points a = x + 2 πi , b = x + 2 πi , 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 π , we have exp( x ) ≤ ρ = exp( u ) ≤ exp( x ), 0 < φ = v ≤ 2 π , which shows that the image of the rectangle is contained in the annulus exp( x ) ≤ ρ ≤ exp( x ). In the opposite direction, for a point ω on the annulus, its polar coordinate ( ρ, φ ) must satisfy exp( x ) ≤ ρ ≤ exp( x ), 0 < φ ≤ 2 π . It is the image of g = log ρ + φi ∈ R under the map ω = e g ....
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This note was uploaded on 02/22/2010 for the course ACM 95A taught by Professor Nilesa.pierce during the Fall '06 term at Caltech.
 Fall '06
 NilesA.Pierce

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