# HW6sol - Z γ sin zdz =-cos π-cos i = 1 1 2 e-1 e 1 = 1 1...

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P ROBLEM SET 6 Partial solution Sec 2.2 p. 2 Note that the function f ( z ) = e z z - 3 is analytic on C \{ 3 } and hence on B (0 , 2) . Therefore by Cauchy’s Formula f (0) = 1 2 πi | z | =2 f ( z ) z - 0 dz Hence | z | =2 e z z ( z - 3) dz = | z | =2 f ( z ) z dz = 2 πif (0) = 2 πi e 0 0 - 3 = - 2 πi 3 . Sec 2.3 p. 6 As computed in Example 6 we have that for z = e cos θ = 1 2 ( z + 1 z ) , sin θ = 1 2 i ( z - 1 z ) , and = dz iz Hence 2 π 0 3 + sin θ + cos θ = | z | =1 2 z 6 z - i ( z 2 - 1) + z 2 + 1 dz iz = | z | =1 - 2 i (1 - i ) z 2 + 6 z + 1 + i dz Factoring the denominator using the quadratic formula we get that (1 - i ) z 2 + 6 z + 1 + i = (1 - i )( z - z 1 )( z - z 2 ) where z 1 , 2 = - 3 ± 7 2 (1 + i ) Note that | z 1 | < 1 and | z 2 | > 1 . Let f ( z ) = 1 z - z 2 . Then, using the Cauchy’s formula 2 π 0 3 + sin θ + cos θ = | z | =1 - 2 i 1 - i f ( z ) z - z 1 dz = - 2 i 1 - i 2 πif ( z 1 ) = 2(1 + i ) π 1 z 1 - z 2 = 2(1 + i ) π 1
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Unformatted text preview: Z γ sin zdz =-cos π-(-cos i ) = 1 + 1 2 ( e-1 + e 1 ) = 1 + 1 2 e + e 2 Sec 2.3 p. 14 Theorem 4 is the Cauchy’s Formula. Let γ be a circle of radius r > centered at z . Parameterizing it by γ ( t ) = re it for ≤ t < 2 π yields f ( z ) = 1 2 πi Z γ f ( ξ ) ξ-z dξ = 1 2 πi Z 2 π f ( z + re it ) re it re it i dt = 1 2 π Z 2 π f ( z + re i t ) dt which is what we wanted to prove....
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