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math185f08-hw7sol

# math185f08-hw7sol - MATH 185 COMPLEX ANALYSIS FALL 2008/09...

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MATH 185: COMPLEX ANALYSIS FALL 2008/09 PROBLEM SET 7 SOLUTIONS For any S C , recall that ¯ S denotes its closure and ∂S denotes its boundary. For any real or complex-valued function f , we will write f k ( z ) := [ f ( z )] k for k N . We will use the shorthand | z - a | = r to denote the circle of radius r centered at a , parameterized by z : [0 , 2 π ] C , z ( t ) = a + re it . 1. Let Ω be a region and D (0 , 1) Ω. Let f : Ω C be analytic. (a) Show that 2 π Z 2 π 0 f ( e ) cos 2 θ 2 = 2 f (0) + f 0 (0) . Solution. By the generalized Cauchy integral formula, f (0) = 1 2 πi Z | z | =1 f ( z ) z dz = 1 2 π Z 2 π 0 f ( e ) dθ, (1.1) f 0 (0) = 1 2 πi Z | z | =1 f ( z ) z 2 dz = 1 2 π Z 2 π 0 f ( e ) e - dθ. (1.2) By Cauchy’s theorem, 0 = 1 2 πi Z | z | =1 f ( z ) dz = 1 2 π Z 2 π 0 f ( e ) e dθ. (1.3) If we take 2 × (1.1) + (1.2) + (1.3), we get 2 f (0) + f 0 (0) = 1 2 π Z 2 π 0 f ( e )(2 + e + e - ) = 1 π Z 2 π 0 f ( e )(1 + cos θ ) = 2 π Z 2 π 0 f ( e ) cos 2 θ 2 as required. (b) Show that for n N , n ! π Z 2 π 0 [Re f ( e )] e - inθ = f ( n ) (0) . Solution. By the generalized Cauchy integral formula, f ( n ) (0) = n ! 2 πi Z | z | =1 f ( z ) z n +1 dz = n ! 2 π Z 2 π 0 f ( e ) e - inθ dθ. Now observe that the given integral may be written as n ! π Z 2 π 0 [Re f ( e )] e - inθ = n ! 2 π Z 2 π 0 [ f ( e ) + f ( e )] e - inθ = f ( n ) (0) + n ! 2 π Z 2 π 0 f ( e ) e - inθ dθ. Date : November 14, 2008 (Version 1.0). 1

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We will get what we need if we could show that the last integral is 0. By Theorem 4.3 , f has a power series representation f ( z ) = X n =0 a n z n that has radius of convergence r > 1 (since D (0 , 1) Ω).
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math185f08-hw7sol - MATH 185 COMPLEX ANALYSIS FALL 2008/09...

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