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Unformatted text preview: Residues and Contour Integration Problems Classify the singularity of f ( z ) at the indicated point. 1. f ( z ) = cot( z ) at z = 0. Ans. Simple pole. Solution. The test for a simple pole at z = 0 is that lim z → z cot( z ) exists and is not 0. We can use L’ Hˆopital’s rule: lim z → z cot( z ) = lim z → z cos( z ) sin( z ) = lim z → cos( z ) z sin( z ) cos( z ) = 1 . Thus the singularity is a simple pole. 2. f ( z ) = 1+cos( z ) ( z π ) 2 at z = π . Ans. Removable. Solution. Power series is the simplest way to do this. We can expand cos( z ) in a Taylor series about z = π . To do so, use the trig identity cos( z ) = cos( z π ). Next, expand 1 cos( z π ) in a power series in z π : 1 + cos( z ) = 1 cos( z π ) = 1 2! ( z π ) 2 1 4! ( z π ) 4 + ··· From this, we get 1 + cos( z ) ( z π ) 2 = ( z π ) 2 ( 1 2 1 4! ( z π ) 2 + ··· ) ( z π ) 2 = 1 2 1 4! ( z π ) 2 + ··· , which is the Laurent series for 1+cos( z ) ( z π ) 2 . Since there are no negative powers in the series, the singularity is removable. 3. f ( z ) = sin(1 /z ). Ans. Essential singularity. 4. f ( z ) = z 2 z z 2 +2 z +1 at z = 1. Ans. Pole of order 2. 5. f ( z ) = z 3 sin( z ) at z = 0. Ans. Pole of order 2. 6. f ( z ) = csc( z ) cot( z ) at z = 0. Ans. Pole of order 2. 1 Find the residue of g ( z ) at the indicated singulatity. 7. g ( z ) = 1 z 2 +1 at z = i . Ans. Res i ( g ) = 1 2 i . Solution. Since g ( z ) = 1 ( z i )( z + i ) , we have that ( z + i ) g ( z ) = 1 z i , which is analytic and nonzero at z = i . Hence, g ( z ) has a simple pole at z = i . The residue is thus Res i ( g ) = lim z → i ( z + i ) g ( z ) = 1 2 i = 1 2 i 8. g ( z ) = e z z 3 at...
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This note was uploaded on 06/17/2011 for the course MATH 132 taught by Professor Grossman during the Spring '08 term at UCLA.
 Spring '08
 Grossman
 Integrals

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