math185f09-hw9sol

# math185f09-hw9sol - MATH 185: COMPLEX ANALYSIS FALL 2009/10...

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: MATH 185: COMPLEX ANALYSIS FALL 2009/10 PROBLEM SET 9 SOLUTIONS 1. Consider the functions defined by g a ( z ) = e iz/ 2- 1 e iz/ 2 + 1 and g b ( z ) = e z/ 2- 1 e z/ 2 + 1 . Show that g a maps the set a := { z C | - 1 &lt; Re z &lt; 1 } to D (0 , 1) while g b maps the set b := { z C | - 1 &lt; Im z &lt; 1 } to D (0 , 1). Hence or otherwise, prove the following. Solution. Note that | g b ( z ) | &lt; 1 e z/ 2- 1 e z/ 2 + 1 &lt; 1 | e z/ 2- 1 | 2 &lt; | e z/ 2 + 1 | 2 [Re( e z/ 2 )- 1] 2 + [Im( e z/ 2 )] 2 &lt; [Re( e z/ 2 ) + 1] 2 + [Im( e z/ 2 )] 2 Re( e z/ 2 )- 2Re( e z/ 2 ) + 1 &lt; Re( e z/ 2 ) + 2Re( e z/ 2 ) + 1 Re( e z/ 2 ) &gt; . But Re( e z/ 2 ) = Re(exp( 2 Re z + i 2 Im z )) = exp( 2 Re z )cos( 2 Im z ) &gt; 0 iff cos( 2 Im z ) &gt; 0 in particular this holds when- 1 &lt; Im z &lt; 1, ie. when z b . To deduce the analogous result for g a and a , just note that g a ( z ) = g b ( iz ) and a = i b . (a) Let f : D (0 , 1) C be an analytic function that satisfies f (0) = 0. Suppose | Re f ( z ) | &lt; 1 for all z D (0 , 1). By considering the function g a f or otherwise, prove that | f (0) | 4 . Solution. Note that g a maps a onto D (0 , 1), g a is analytic on a , and g a (0) = 0. The composition F = g a f : D (0 , 1) C is analytic on D (0 , 1); F (0) = g a ( f (0)) = g a (0) = 0; for any z D (0 , 1), | Re f ( z ) | &lt; 1 and so f ( z ) a and so g a ( f ( z )) D (0 , 1), ie. | F ( z ) | = | g a ( f ( z )) | &lt; 1 for all z D (0 , 1). By Schwarzs Lemma, we have | F (0) | 1. Since F ( z ) = g a ( f ( z )) f ( z ) = ie if ( z ) / 2 f ( z ) ( e if ( z ) / 2 + 1) 2 and f (0) = 0, we get | f (0) | 4 . Date : November 30, 2009 (Version 1.0). 1 (b) Let S be the set of functions defined by S = { f : b C | f analytic, | f | &lt; 1 on b , and f (0) = 0 } . By considering the function f g- 1 b or otherwise, Prove that sup f S | f (1) | = e / 2- 1 e / 2 + 1 . Solution. Note that g b maps b into D (0 , 1), g b is analytic on b , and g b (0) = 0. Fur- thermore, note that g b is injective on b g b ( z ) = g b ( w ) iff ( e z/ 2- 1)( e w/ 2 + 1) = ( e z/ 2 + 1)( e w/ 2- 1) iff e ( z- w ) / 2 = 1 iff Re( z- w ) = 0 and Im( z- w ) is an integer multiple of 4; so when z,w b , this is only possible if z = w . Hence g b is an invertible map and g- 1 b maps D (0 , 1) onto b , g- 1 b is analytic on D (0 , 1), and g- 1 b (0) = 0. Given any f S , the composition F = f g- 1 b : D (0 , 1) C is analytic on D (0 , 1); F (0) = f ( g- 1 b (0)) = f (0) = 0; for any z D (0 , 1), g- 1 b ( z ) b and so f ( g- 1 b ( z )) D (0 , 1), ie....
View Full Document

## This note was uploaded on 02/17/2010 for the course MATH 185 taught by Professor Lim during the Fall '07 term at University of California, Berkeley.

### Page1 / 7

math185f09-hw9sol - MATH 185: COMPLEX ANALYSIS FALL 2009/10...

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

View Full Document
Ask a homework question - tutors are online