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math185f09-hw8sol

math185f09-hw8sol - MATH 185 COMPLEX ANALYSIS FALL 2009/10...

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Unformatted text preview: MATH 185: COMPLEX ANALYSIS FALL 2009/10 PROBLEM SET 8 SOLUTIONS 1. Let Ω be a region containing D (0 , 1). Let f : Ω → C be analytic. (a) Let M > 0 be a constant. Suppose | f ( z ) | ≥ M for all z ∈ ∂D (0 , 1) and | f (0) | < M . Show that f has at least one zero in D (0 , 1). Solution. Suppose not and f has no zeros in D (0 , 1). Then by the minimum modulus theorem (or Corollary 4.15 ), M ≤ min z ∈ ∂D (0 , 1) | f ( z ) | = min z ∈ D (0 , 1) | f ( z ) | ≤ | f (0) | < M, a contradiction. So f must have at least one zero in D (0 , 1). (b) Suppose | f ( z 2 ) | ≥ | f ( z ) | for all z ∈ D (0 , 1). Show that f is constant on D (0 , 1). Solution. Let 0 < r < 1. By the compactness of ∂D (0 ,r 2 ), there exists z ∈ ∂D (0 ,r 2 ) such that | f ( z ) | = max z ∈ ∂D (0 ,r 2 ) | f ( z ) | . By the maximum modulus theorem, | f ( z ) | = max z ∈ D (0 ,r 2 ) | f ( z ) | . But since | f ( z 2 ) | ≥ | f ( z ) | , | f ( z ) | = max z ∈ D (0 ,r 2 ) | f ( z ) | = max z ∈ D (0 ,r ) | f ( z 2 ) | ≥ max z ∈ D (0 ,r ) | f ( z ) | . (1.1) Note that r 2 < r and so z ∈ ∂D (0 ,r 2 ) ⊂ D (0 ,r ). So | f ( z ) | ≤ max z ∈ D (0 ,r ) | f ( z ) |...
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math185f09-hw8sol - MATH 185 COMPLEX ANALYSIS FALL 2009/10...

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