math185f08-hw8sol - MATH 185: COMPLEX ANALYSIS FALL 2008/09...

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FALL 2008/09 PROBLEM SET 8 For z C × , recall that the principle argument of z , denoted Arg( z ), is the unique θ [ - π,π ) such that z = | z | e . By convention Arg(0) = 0. 1. Let S denote the sector given by { z C | - π/ 4 < Arg( z ) < π/ 4 } . Let f : ¯ S C be a continuous function such that f is analytic on S . Suppose (i) | f ( z ) | ≤ 1 for all z ∂S ; (ii) | f ( x + iy ) | ≤ e x for all x + iy S . Prove that | f ( z ) | ≤ 1 for all z S . Solution. Let ε > 0. Consider the function F ( z ) = e - εz f ( z ). Then F is also continuous on ¯ S and analytic in S . By (i), | F ( z ) | = e - εx | f ( z ) | ≤ 1 for z ∂S . By (ii), lim z S, | z |→∞ | F ( z ) | ≤ lim x →∞ e - εx e x = lim x →∞ e - εx + x = 0 . Hence there exists R > 0 such that | F ( z ) | ≤ 1 for all z ¯ S , | z | ≥ R , i.e. max z ¯ S ( C \ D (0 ,R )) | F ( z ) | ≤ 1 . Since F is continuous, it attains its maximum on the compact set ¯ S D (0 ,R ) and so we may apply the maximum modulus theorem to F to conclude that max z ¯ S D (0 ,R ) | F ( z ) | = max z ( S D (0 ,R )) | F ( z ) | = max ( max z ∂S D (0 ,R ) | F ( z ) | , max z S ∂D (0 ,R ) | F ( z ) | ) 1 . Hence max z ¯ S | F ( z ) | ≤ 1 and so for all z ¯ S , | f ( z ) | ≤ | e εz | = e ε Re( z ) . Since ε > 0 is arbitrary, for each fixed z ¯ S , we may take right limit to get | f ( z ) | ≤ lim ε 0 + e ε Re( z ) = 1 . Hence | f ( z ) | ≤ 1 for all z ¯ S , as required. Date
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math185f08-hw8sol - MATH 185: COMPLEX ANALYSIS FALL 2008/09...

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