This preview shows pages 1–2. Sign up to view the full content.
This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
Unformatted text preview: MAT 461 Assignment E  Solutions Posted November 30th 2007 Evaluate the following integrals, where C is the circle  z  = 5 in the positive sense. (1) Z C sin z ( z 2 1)( z + 10) dz The function has singularities at z = 1 , 1 , 10 of which 1 , 1 are inside the contour C . We must determine the residues at z = 1 , 1. C is simple and positive, so the answer will be 2 i times the sum of the residues at singularities inside. I will use the f over q method. f ( z ) = sin z q ( z ) = ( z 2 1)( z + 10) q ( z ) = 2 z ( z + 10) + 1( z 2 1) = 3 z 2 + 20 z 1 f (1) = sin1 6 = 0 q (1) = 3 + 20 1 = 22 6 = 0 Res z =1 = f (1) q (1) = sin1 22 f ( 1) = sin( 1) = sin1 6 = 0 q ( 1) = 3 20 1 = 18 6 = 0 Res z = 1 = f ( 1) q ( 1) = sin1 18 Z C sin z ( z 2 1)( z + 10) dz = 2 i ( sin1 22 + sin1 18 ) (2) Z C e 1 z (4 z 3 3) z ( z 3 + 1) dz In making this question, my plan was to make an integral so horrible to do in the standard way that you would be forced to think outside the box and use the Residue at method. The singularities for this function are at z = 0 , 1 ,e i 2 3 ,e i 2 3 . These are all inside the curve C . So we are able to use the the residue at infinity method (which require all singularities be inside the contour)....
View
Full
Document
 Fall '07
 Ibrahim
 Integrals

Click to edit the document details