math185-hw6sol - MATH 185 COMPLEX ANALYSIS FALL 2007/08...

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MATH 185: COMPLEX ANALYSIS FALL 2007/08 PROBLEM SET 6 SOLUTIONS You may use without proof any results that had been proved in the lectures. 1. Let Ω C be a region. Let f be analytic on Ω and let z 0 Ω. Suppose f 0 ( z 0 ) 6 = 0. Show that there is an r > 0 such that Z Γ f 0 ( z 0 ) f ( z ) - f ( z 0 ) dz = 2 πi where Γ = ∂D ( z 0 , r ). Solution. We know that there is an R > 0 such that f has a Taylor series expansion about z 0 . So f ( z ) = f ( z 0 ) + a 1 ( z - z 0 ) + X n =2 a n ( z - z 0 ) n holds for all z D ( z 0 , R ). Now since a 1 = f 0 ( z 0 ) 6 = 0 and since f 0 is continuous at z 0 , there is an δ > 0 such that f ( z ) - f ( z 0 ) 6 = 0 for all z D ( z 0 , δ ) \{ z 0 } (if not, we can find a sequence z n z 0 , z n 6 = z 0 , such that f ( z n ) - f ( z 0 ) = 0 for all n N — this will imply that 0 = lim n →∞ ( f ( z n ) - f ( z 0 )) / ( z n - z 0 ) = f 0 ( z 0 ), a contradiction). Let r = min { R, δ } and let the function g : D ( z 0 , r ) C be defined by g ( z ) = f ( z ) - f ( z 0 ) z - z 0 z 6 = z 0 , f 0 ( z 0 ) z = z 0 . Now observe that g is analytic in D ( z 0 , r ) by a result in the lectures. Furthermore, g is non-zero on D ( z 0 , r ). Hence the function h : D ( z 0 , r ) C defined by h ( z ) = 1 g ( z ) is analytic on D ( z 0 , r ). Cauchy’s integral formula applied to h yields 1 2 πi Z Γ h ( z ) z - z 0 dz = h ( z 0 ) but since h ( z ) = z - z 0 f ( z ) - f ( z 0 ) z 6 = z 0 , 1 f 0 ( z 0 ) z = z 0 , we get 1 2 πi Z Γ 1 f ( z ) - f ( z 0 ) dz = 1 f 0 ( z 0 ) and thus Z Γ f 0 ( z 0 ) f ( z ) - f ( z 0 ) dz = 2 πi as required. Date : November 1, 2007 (Version 1.3). 1
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2. Let M > 0 and α > 0, α not necessarily an integer. Suppose f is entire and | f ( z ) | ≤ M | z | α for all z C . Show that f is a polynomial. Solution. If α < 1, then 1 - α > 0 and so lim | z |→∞ | f ( z ) | | z | lim | z |→∞ M | z | 1 - α = 0 and f is constant on C by Homework 4 , Problem 5 (a). Now suppose α 1. Since f is entire, f
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