hw7 - Mathematics 185 Intro to Complex Analysis Fall 2009...

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Mathematics 185 – Intro to Complex Analysis Fall 2009 – M. Christ Solutions to Selecta from Problem Set 7 2.5.3 Give an example of a bounded analytic function f on a domain A , such that f is continuous on the closure of A , yet sup z A | f ( z ) | > sup w bd A | f ( w ) | . Solution. Let A = { z : Re( z ) > 0 } and consider f ( z ) = e z . Then bd( A ) is the imaginary axis, and | f ( w ) | = 1 for all w bd( A ), yet f is not even bounded on A . 2.5.4(a) Let R > 0 and suppose that | z 0 | < R . Let D = D (0 ,R ) and D 0 = D (0 , 1). Define T ( z ) = R ( z - z 0 ) R 2 - ¯ z 0 z . Show that T maps D bijectively to D 0 , and that T ( z 0 ) = 0. Solution. Certainly T ( z 0 ) = 0. First suppose that R = 1, to simplify the formulas; see below for the general case. Note that the denominator does not vanish for z in question: | 1 - ¯ z 0 z | ≥ 1 - | z 0 | · | z | ≥ 1 - | z 0 | > 0 so long as | z | ≤ 1. Now if | z | = 1 then since z - 1 = ¯ z , | z - z 0 | | 1 - ¯ z 0 z | = | z - z 0 | | z | · | ¯ z - ¯ z 0 z | = | z | - 1 | z - z 0 | | ¯ z - ¯ z 0 z | = 1 since | z - z 0 | = | ¯ z - ¯ z 0 | . In other words, | T ( z ) | ≤ 1 for all z in the closure of D . By the maximum principle, | T ( z ) | ≤ 1 whenever | z | ≤ 1. The quickest way to show that T is a bijection is to find its inverse algebraically: Solving the equation T ( z ) = w gives z = ˜ T ( w ) = w + z 0 1+ w ¯ z 0 . ˜ T is of the same form as T , but with z 0 replaced by - z 0 ; so what we have proved above shows that ˜ ( D 0 ) D 0 . Thus T : D 0 D 0 is a bijection, with inverse ˜ T . Now for the general case, just write
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This note was uploaded on 01/19/2010 for the course MATH 113 taught by Professor Ogus during the Fall '08 term at University of California, Berkeley.

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hw7 - Mathematics 185 Intro to Complex Analysis Fall 2009...

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