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Unformatted text preview: Page 711, problem 7: ( 29 ( 29 ( 29 ( 29 ( 29 4 4 4 4 4 4 1 1 ) 3 sin( 1 ) 3 sin( ) ( i z i z z z z z z z f + + = = implies fourth order poles at . , 1 i z = Let z=1/w + + = = 5 3 ! 5 3 ! 3 3 3 3 sin ) 3 sin( w w w w z implies that f(z) has an essential singularity at infinity. Page 711, problem 8: ( 29 ( 29 3 2 1 8 1 2 1 4 ) ( + = z z z z f implies that f(z) has a third order pole at z=1. f(z)>0 as z approaches infinity, so there is no singularity at infinity. Page 717, problem 9: ( 29 ( 29 ( 29 2 2 2 2 1 1 1 1 1 z z z = + so there are second order poles at 1 z = . Residue at z=1: ( 29 ( 29 2 3 1 1 1 2 1 lim lim 4 1 1 z z d dz z z  = =  + + Residue at z=1: ( 29 ( 29 2 3 1 1 1 2 1 lim lim 4 1 1 z z d dz z z   = =  Page 717, problem 14: The singularity at z=0 is inside the contour....
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This note was uploaded on 09/05/2011 for the course EGM 4313 taught by Professor Mei during the Spring '08 term at University of Florida.
 Spring '08
 MEI

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