midsol

# midsol - ACM 95a/100a Midterm Solutions Prepared by Zhiyi...

This preview shows pages 1–3. Sign up to view the full content.

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
This is the end of the preview. Sign up to access the rest of the document.

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 g-strips of width 2 π to the whole ω-plane (except for the origin). The strip R = { g : 0 < I ( g ) ≤ 2 π,-∞ < R ( g ) < ∞} (Figure 2) maps one-to-one 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 ....
View Full Document

{[ snackBarMessage ]}

### Page1 / 12

midsol - ACM 95a/100a Midterm Solutions Prepared by Zhiyi...

This preview shows document pages 1 - 3. Sign up to view the full document.

View Full Document
Ask a homework question - tutors are online