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Unformatted text preview: Key to Homework 8, Thanks to Da Zheng for the textfile November 18, 2011 1. Prove that csc z = 1 z + 2 z + X n =1 ( 1) n 1 z 2 n 2 2 , z 6 = 0 , , 2 , Proof. We consider the following auxiliary function, where z 6 = 0 , , 2 , f ( ) = csc  1 (  z ) Then, it is easy to check that csc  1 has a removable singularity at = 0, so we will regard it holomorphic at = 0. Thus, the function f has poles at = n , n Z , each of order one.Here, notice that = 0 is also a removable singularity for f . Next, choosing the following sequence of counterclockwise square contour n , centered at the origin with vertices ( n + 1 2 )( 1 i ) . Hence, integrating f ( ) over contour n and applying the residues theorem, we obtain 1 2 i I n f ( ) d = csc z 1 z z + n X k =1 ( 1) k 1 k ( k z ) + n X k =1 ( 1) k 1 k ( k + z ) ( * ) Then, for enough large n (i.e. n contains z ), we evaluate the integral on the left hand side as follows: we first claim that csc z is bounded on the boundary of the square n ( please see this argument carefully ): Indeed, let z = x + iy , for y > 2 ,  csc z  = 2 i e iz e iz 2 e y e y 2 e / 2 e / 2 1 . Similarly, for y < 2 ,  csc z  1. In other words,  csc z  1 on two horizontal lines of the square and also on the part of two vertical lines with  y  > 2 . Hence, 1 it only remains to see the points on the part of two vertical lines with  y  2 . Notice that on the line segment with ends 2 2 i, 2 + 2 i , the function sin z has no zeros, so  csc z  M 1 on the line segment. Since  csc( z + )  =  csc z  , so  csc z  M 1 on the line segment with ends n + 2 2 i,n + 2 + 2 i . Hence  csc z  M on n . Thus I n f ( ) d = I n csc  1 (  z ) d I n csc  1 (  z )  d  I n  csc   (  z )  +  1   (  z )   d  2 M ( n + 1 / 2)   z  + 2 ( n + 1 / 2) (( n + 1 / 2)   z  ) , as n + So, the expression (*) becomes 0 = csc z 1 z z + X k =1 ( 1) k 1 k ( k z ) + X k =1 ( 1) k 1 k ( k + z ) Simply the above and we obtain our desired equality csc z = 1 z + 2 z + X n =1 ( 1) n 1 z 2 n 2 2 , z 6 = 0 , , 2 , 2 2. Find the expression of + X 1 n 2 + a 2 where a > 0 is not a integer....
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This note was uploaded on 02/21/2012 for the course MATH 4377 taught by Professor Staff during the Summer '08 term at University of Houston.
 Summer '08
 Staff

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