MA 513 HW 8

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

This preview shows pages 1–2. Sign up to view the full content.

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:

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

Page1 / 2

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

This preview shows document pages 1 - 2. Sign up to view the full document.

View Full Document
Ask a homework question - tutors are online