A series for sin az / sin pi z in complex analysis

[Gunni]

Homework Statement

Show that

[tex] \frac{\sin (az)}{\sin (\pi z)} = \frac{2}{\pi} \sum_{n=1}^{+\infty} (-1)^n \frac{n \sin (an)}{z^2 - n^2} [/tex]

for all a such that [tex] - \pi < a < \pi [/tex]

Homework Equations

None really, we have similar expansions for [tex]\pi cot (\pi z)[/tex] and [tex] \pi / \sin (\pi z) [/tex], this is an excersize in using Mittag-Leffler's theorem.

The Attempt at a Solution

My problem is that I can't show the series is uniformly convergent on every compact subset of C. Once that's done I've got a solution for every rational multiple of pi, which I think can be extended to all real a with a continuity argument. Any thoughts on the convergence problem? It's driving me mad.

 

[lippyka]

I should also note that I have the result for a=0, which gives \frac{1}{\sin (\pi z)} = \frac{2}{\pi} \sum_{n=1}^{+\infty} \frac{(-1)^n}{z^2 - n^2} which turns out to be uniformly convergent on every compact subset of C.