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# 814hw5sol - Math 814 HW 5 p 87 6 7 p 96 8a 10 11 p 110...

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Math 814 HW 5 December 11, 2007 p. 87: 6, 7 p. 96: 8a, 10, 11 p. 110: 1bchi, 5,13. p.87, no. 6. Let f be analytic on D = B (0 , 1) and suppose | f ( z ) | ≤ 1 on D . Show that | f (0) | ≤ 1 . Proof: Let 0 < r < 1 and let γ r ( t ) = re it for 0 t 2 π . By the Cauchy Integral Formula, we have f (0) = 1 2 πi γ r f ( w ) w 2 dw, so | f (0 | ≤ 1 2 π · 1 r 2 · 2 πr = 1 r . Taking the limit as r 1 , we have | f (0) | ≤ 1 . p.87, no. 7. Let γ ( t ) = 1 + e it for 0 t 2 π and let n N . Find γ z z - 1 n dz. Apply C.I.F. to f ( z ) = z n , to get γ z z - 1 n dz = 2 πi ( n - 1)! · f n - 1 (1) = 2 nπi. 1

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You can also do this without C.I.F., by computing directly: γ z z - 1 n dz = 2 π 0 1 + e it e it n · ie it dt = i 2 π 0 ( 1 + e - it ) n · e it dt = i 2 π 0 1 + ne - it + n 2 e - 2 it + · · · · e it dt = 2 nπi, since 2 π 0 e kit dt = 0 for k a nonzero integer. p.96, no. 8a. We must integrate ( z - a ) - 1 and ( z - b ) - 1 over the path γ , which can be written as a sum of six paths, two of which are closed and have a, b in their -components, hence have zero integral and two pairs of non-closed paths. One pair starts at the leftmost crossing point, each goes around a in opposite directions, and they meet at the middle crossing point. The other one pair starts at the middle crossing point, each goes around b
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