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

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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 π 2 , z 6 = 0 , ± π, ± 2 π, · · · Proof. We consider the following auxiliary function, where z 6 = 0 , ± π, ± 2 π, · · · f ( ζ ) = csc ζ - 1 ζ ζ ( ζ - z ) Then, it is easy to check that csc ζ - 1 ζ has a removable singularity at ζ = 0, so we will regard it holomorphic at ζ = 0. Thus, the function f has poles at ζ = , n Z , each of order one.Here, notice that ζ = 0 is also a removable singularity for f . Next, choosing the following sequence of counterclockwise square contour γ n , centered at the origin with vertices ( n + 1 2 )( ± 1 ± i ) π . Hence, integrating f ( ζ ) over contour γ n and applying the residue’s theorem, we obtain 1 2 πi I γ n f ( ζ ) = csc z - 1 z z + n X k =1 ( - 1) k 1 ( - z ) + n X k =1 ( - 1) k 1 ( + z ) ( * ) Then, for enough large n (i.e. γ 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 γ n ( please see this argument carefully ): Indeed, let z = x + iy , for y > π 2 , | csc z | = 2 i e iz - e - iz 2 e y - e - y 2 e π/ 2 - e - π/ 2 1 . Similarly, for y < - π 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 | > π 2 . Hence, 1

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it only remains to see the points on the part of two vertical lines with | y | ≤ π 2 . Notice that on the line segment with ends π 2 - π 2 i, π 2 + π 2 i , the function sin z has no zeros, so | csc z | ≤ M 1 on the line segment. Since | csc( z + π ) | = | csc z | , so | csc z | ≤ M 1 on the line segment with ends n + π 2 - π 2 i, n + π 2 + π 2 i . Hence | csc z | ≤ M on γ n . Thus I γ n f ( ζ ) = I γ n csc ζ - 1 ζ ζ ( ζ - z ) I γ n csc ζ - 1 ζ ζ ( ζ - z ) | | I γ n | csc ζ | | ζ ( ζ - z ) | + | 1 ζ | | ζ ( ζ - z ) | | | 2 πM ( n + 1 / 2) π - | z | + 2 π ( n + 1 / 2) π (( n + 1 / 2) π - | z | ) 0 , as n + So, the expression (*) becomes 0 = csc z - 1 z z + X k =1 ( - 1) k 1 ( - z ) + X k =1 ( - 1) k 1 ( + 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 π 2 , z 6 = 0 , ± π, ± 2 π, · · · 2 2. Find the expression of + X -∞ 1 n 2 + a 2 where a > 0 is not a integer.
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HW8s - Key to Homework 8 Thanks to Da Zheng for the text-le...

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