SOLUTION TO HOMEWORK ON PAGE 121


6. Let f(z) = pi cot(pi z)/(z+a)^2. Then, f has a pole of order 2 at z=-a
   and simple poles at integers z=k, k an integer.

   Res(f,-a) = -pi^2 csc^(2)(pi a)
   Res(f,k) = (k+a)^(-2) ,  using the LaHopital's rule.

   Thus,

   int_{gamma}f = 2pi{ -pi^2 csc^(2)(pi z) + sum_{-n}^{n}(k+a)^(-2).

   Before we prove the claim that int_{gamma}f -> 0 as n->infty, note
   that the final result would be a trivial consequence of this claim. So,
   it only remains to prove the claim. The hard part of the proof is to 
   show that for sufficiently large k,|cot(pi z)| < 2 (any fixed constant
   would do).  So let z=x+iy and write

   cot(pi z) = cos(pi z) / sin(pi z)

   = i {e^(i pi z) + e^(-i pi z)} / {e^(i pi z) - e^(-i pi z)}
   
   = i{e^{2i pi z) + 1} / {e^{2i pi z) - 1}

   Now consider the term in the denomenator.

   |e^(2i pi z) - 1| \geq 1 - e^(-2 pi y) \geq 1 by choosing y large enough.

   And similarly the term in the numerator,

   |e^(2i pi z) + 1| \leq 1 + 2^(-2 pi y) \leq 1+1 = 2 by choosing y large

   enough.  Hence by choosing n (and hence z and thus y) large enough we
   have that |cot(pi z)| \leq 2. The rest of the proof is straight forward.

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7. In problem number 6 let a=1/2 to get

   (pi^2)/4 = sum_{-infty}^{infty} (2n+1)^(-2)

   Now split this sum into two, one from 0 to infinity and one from -1 to
   -infinity. In the sum from -1 to -infty replace n by -n-1, it follows
   that

   sum_{-1}^{-infty} (2n+1)^(-2} = sum_{0}^{infty} (2n+1)^(-2).

   From this identity the result will follow immediately.