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Unformatted text preview: (1) Show that f ( z ) = 1 z + ∑ ∞ n =1 2 z z 2 n 2 is an analytic function on C \ N . Show that the singular points are all simple (= order 1) poles and determine their residues. Show that f is 1periodic and odd. Finally show that f ( z ) = π cot( πz ). Solution: Note first that for fixed z 6∈ Z the series ∑ ∞ n =1 2 z z 2 n 2 con verges, as we can find a bound  2 z z 2 n 2  ≤ c/n 2 for some constant c > 0. So at least f ( z ) is welldefined in each point. Then on a compact subset K of C \ Z we note there is a constant M such that  z  < M if z ∈ K . Thus we can find a bound for all z ∈ K and n > M + 1 2 z z 2 n 2 = 2 z ( z n )( z + n ) ≤ 2 M n 2 M 2 ≤ 2 M 1 M 2 / ( M + 1) 2 1 n 2 , Define f j ( z ) = 1 z + ∑ j n =1 2 z z 2 n 2 , then clearly f j is a holomorphic function on C \ Z , and f j → f as j → ∞ uniformly on K (as the tail ends can be uniformly (for z ∈ K ) bounded by  f ( z ) f j ( z )  ≤ X n ≥ j +1  2 z z 2 n 2  ≤ 2 M 1 M 2 / ( M + 1) 2 X n ≥ j +1 1 n 2 which converges to 0 as j → ∞ .) Hence we conclude that f is holomorphic on C \ Z . To prove that n ∈ Z is a simple pole for f (with residue 1) we will show that f ( z ) 1 z n has a removable singularity at z = n . Indeed, write f j ( z ) = 1 z + j X m =1 2 z z 2 m 2 = 1 z + j X m =1 1 z m + 1 z + m = j X m = j 1 z m , and we see that f j has a simple pole at z ∈ { j, j +1 ,...,j } with residue 1. Thus ( z n ) f j has a removable singularity at z = n for all j , and hence ( z n ) f has a removable singularity at z = n and f has at most a simple pole at z = n . Moreover for j > n we see that lim z → n ( z n ) f j ( z ) = 1, while f j f is a holomorphic function at z = n (prove as before), so the residue of f at z = n equals the residue of f j at z = n equals 1. In particular f is not regular at z = n , it has a pole, and by the previous argument this is simple....
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 Spring '09
 Periodicity, 2m, Pole, fJ

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