HW8s - Key to Homework 8, Thanks to Da Zheng for the...

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Unformatted text preview: Key to Homework 8, Thanks to Da Zheng for the text-file 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.

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HW8s - Key to Homework 8, Thanks to Da Zheng for the...

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