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EGM4313.HW11 - Page 711 problem 7 f z = sin(3 z(z 4 1 4...

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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. ( 29 ( 29 3 3 4 4 3 sin 1 3! 6 z z z z z z z π π π π π = - +⋅⋅⋅ = - +⋅⋅⋅ ( 29 ( 29 3 4 4 sin 2 6 3 C z dz i i z π π π π = - = - Ñ (You could also find the residue by using the formula for a third order pole.)
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