MA 513 HW 8

MA 513 HW 8 - MA 513 Homework 8 Solutions p.255#5...

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MA 513 Homework 8 Solutions p.255: #5; p.267: #1,5,6; p. 275: #4,7; p.286: #2. p. 255: #5: Within the contour C N the function f ( z ) = 1 z 2 sin z had singularities z = 0 ,z = ± nπ, n = 1 , 2 ,...N. At z = 0 , sin z = z - z 3 / 6 + ..., so that f ( z ) has a pole of order 3; the residue is 1 / 6 . At z = ± nπ, f ( z ) has a simple pole, with residue ( - 1) n n 2 π 2 . Therefore, i C N 1 z 2 sin z dz = 2 πi p 1 6 + 2 N s 1 ( - 1) n n 2 π 2 P . (The 2 comes from poles at z = ± with the same residue.) Letting N → ∞ , and observing that the integral approaches zero, the sum in the parentheses must be zero in the limit. Rearranging, we get N s 1 ( - 1) n n 2 = - π 2 12 . p. 267: #1: 1 z 2 + 1 has a simple pole at z = i inside the semicircle radius R > 1 in the upper half plane. Integral comes from the residue of 1 / 2 i and letting R → ∞ . #5:
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MA 513 HW 8 - MA 513 Homework 8 Solutions p.255#5...

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