Showing that $-4\pi^2\sum_{m=1}^\infty me^{2\pi i m\tau}=\frac{\pi^2}{\sin^2(\pi \tau)}$

Question

I'm doing Problem 7b in Chapter 4 of Stein and Shakarchi's "Complex Analysis" for homework. I want to show that if $\tau$ is a complex number with $\mathrm{Im}(\tau)>0$, then $$\sum_{n=-\infty}^\infty \frac{1}{(\tau+n)^2}=\frac{\pi^2}{\sin^2(\pi \tau)}$$ by using the identity $$\sum_{n=-\infty}^\infty \frac{1}{(\tau+n)^2}=-4\pi^2\sum_{m=1}^\infty me^{2\pi i m\tau}$$ and I have no idea how to do this. It's not even clear to me that the sum on the right converges. This is in the chapter on the Fourier transform, so it seems like I would want to use the Poisson summation formula, but that was already used to derive the identity. Another thing that seemed possibly useful was the fact that $$\frac{1}{e^{2\pi i z}-1}=e^{-2\pi i z}\sum_{n=0}^\infty e^{-2\pi i nz}$$ but I can't use that since the exponential is multiplied by $m$ in the sum. Does anyone have any hints for what I could try?

Answer 1

Hint

Take the derivative of your last expression according to $z$, to reach the previous sum involving $me^{2\pi i m \tau}$ on the right hand side, and on the LHS a expression proportional to $\sin^{-2}(\pi z)$.