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

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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 < Re z < 1 } to D (0 , 1) while g b maps the set b := { z C | - 1 < Im z < 1 } to D (0 , 1). Hence or otherwise, prove the following. Solution. Note that | g b ( z ) | < 1 e z/ 2- 1 e z/ 2 + 1 < 1 | e z/ 2- 1 | 2 < | e z/ 2 + 1 | 2 [Re( e z/ 2 )- 1] 2 + [Im( e z/ 2 )] 2 < [Re( e z/ 2 ) + 1] 2 + [Im( e z/ 2 )] 2 Re( e z/ 2 )- 2Re( e z/ 2 ) + 1 < Re( e z/ 2 ) + 2Re( e z/ 2 ) + 1 Re( e z/ 2 ) > . But Re( e z/ 2 ) = Re(exp( 2 Re z + i 2 Im z )) = exp( 2 Re z )cos( 2 Im z ) > 0 iff cos( 2 Im z ) > 0 in particular this holds when- 1 < Im z < 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 ) | < 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 ) | < 1 and so f ( z ) a and so g a ( f ( z )) D (0 , 1), ie. | F ( z ) | = | g a ( f ( z )) | < 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 | < 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....
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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.

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

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