# HW9 - n-1 | | a n | z | ≤ a R n + a R n-1 + · · · + a...

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MATH 185 SOLUTIONS OF HW IX (1) (50. 1)(30pts) Let z o C . Then for any R > 0, by Cauchy’s inequality, we have | f 00 ( z 0 ) | ≤ 2 M R R 2 where M R is the maximum value of | f ( z ) | for all z C R and C R is the circle centered at z 0 and with radius R . Since | f ( z ) | ≤ A | z | ≤ A ( | z 0 | + R ), M R A ( | z 0 | + R ), so we have | f 00 ( z 0 ) | ≤ 2 A ( | z 0 | + R ) R 2 Since we consider R as an arbitrary number, so the right hand side of above inequality gets smaller as we take a larger R . i.e. it goes to zero when R goes to inﬁnity. Hence, | f 00 ( z 0 ) | = 0, i.e. f 00 ( z 0 ) = 0 for all z o C . Thus we have f 0 ( z 0 ) is constant and hence f ( z ) = az + b for some a,b C . By substituting z = 0 on the given inequality, we have | f (0) | ≤ A | 0 | = 0, i.e. f (0) = 0. Thus f ( z ) = az for some a C . (2) (50. 3)(30pts) Let a = max {| a 0 | / | a n | , | a 1 | / | a n | , · · · , | a n - 1 | / | a n |} . Let take R > max { na, 1 } . Then, | a 0 | | a n || z n | + | a 1 | | a n || z n - 1 | + · · · + | a
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Unformatted text preview: n-1 | | a n | z | ≤ a R n + a R n-1 + · · · + a R ≤ na R < 1 for all | z | ≥ R . Thus, we have | P ( z ) | < 2 | a n || z n | whenever | z | ≥ R . (3) (50. 7)(40pts) Let g ( z ) = e f ( z ) , then g ( z ) is analytic on a closed bounded region R and not a constant on R . Hence the maximum and minimum modulus of g ( z ) must be taken on the boundary of R , so | g ( z ) | = e u ( x,y ) has its minimum value on the boundary of R . Since z = e x is a monotonic increasing function of x ∈ R , u ( x,y ) has its minimum value on the boundary of R . 1...
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## This note was uploaded on 09/01/2010 for the course MATH 185 taught by Professor Lim during the Fall '07 term at Berkeley.

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