Expression for Jacobi theta function derivative

Question

Please explain how to get $$\frac{\vartheta'_{1}(z, q)}{\vartheta_{1}(z, q)} = \cot z + 4\sum_{n = 1}^{\infty}\frac{q^{2n}}{1 - q^{2n}}\sin 2nz$$ where $\vartheta_{1}(z, q)$ is one of the Jacobi theta functions. Thanks!

Answer 1

The infinite product representation for $\vartheta_{1}(z, q)$ given by $$\vartheta_{1}(z, q) = 2q^{1/4}\sin z\prod_{n = 1}^{\infty}(1 - q^{2n})(1 - 2q^{2n}\cos 2z + q^{4n})\tag{1}$$ I hope you are familiar with the above result. Taking logs we get \begin{align}\log\vartheta_{1}(z, q) &= \log(2q^{1/4}) + \log\sin z + \log\prod_{n = 1}^{\infty}(1 - q^{2n})\notag\\ &\,\,\,\,\,\,\,\, + \sum_{n = 1}^{\infty}\log(1 - 2q^{2n}\cos 2z + q^{4n})\tag{2} \end{align} and differentiating with respect to $z$ we get \begin{align} \frac{\vartheta'_{1}(z, q)}{\vartheta_{1}(z, q)} &= \cot z + 4\sum_{n = 1}^{\infty}\frac{q^{2n}\sin 2z}{1 - 2q^{2n}\cos 2z + q^{4n}}\notag\\ &= \cot z + \frac{4}{2i}\sum_{n = 1}^{\infty}\frac{q^{2n}(e^{2iz} - e^{-2iz})}{(1 - q^{2n}e^{2iz})(1 - q^{2n}e^{-2iz})}\notag\\ &= \cot z + \frac{4}{2i}\sum_{n = 1}^{\infty}q^{2n}\left(\frac{e^{2iz}}{1 - q^{2n}e^{2iz}} - \frac{e^{-2iz}}{1 - q^{2n}e^{-2iz}}\right)\notag\\ &= \cot z + \frac{4}{2i}\left(\sum_{n = 1}^{\infty}\frac{q^{2n}e^{2iz}}{1 - q^{2n}e^{2iz}} - \sum_{n = 1}^{\infty}\frac{q^{2n}e^{-2iz}}{1 - q^{2n}e^{-2iz}}\right)\notag\\ &= \cot z + \frac{4}{2i}\left(\sum_{n = 1}^{\infty}\sum_{m = 1}^{\infty}\{q^{2n}e^{2iz}\}^{m} - \sum_{n = 1}^{\infty}\sum_{m = 1}^{\infty}\{q^{2n}e^{-2iz}\}^{m}\right)\notag\\ &= \cot z + \frac{4}{2i}\left(\sum_{n = 1}^{\infty}\sum_{m = 1}^{\infty}q^{2m}e^{2imz}q^{2m(n - 1)}\right. \notag\\ &\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, \left. - \sum_{n = 1}^{\infty}\sum_{m = 1}^{\infty}q^{2m}e^{-2imz}q^{2m(n - 1)}\right)\notag\\ &= \cot z + \frac{4}{2i}\left(\sum_{m = 1}^{\infty}q^{2m}e^{2imz}\sum_{n = 1}^{\infty}q^{2m(n - 1)}\right.\notag\\ &\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, \left. - \sum_{m = 1}^{\infty}q^{2m}e^{-2imz}\sum_{n = 1}^{\infty}q^{2m(n - 1)}\right)\notag\\ &= \cot z + \frac{4}{2i}\left(\sum_{m = 1}^{\infty}\frac{q^{2m}e^{2imz}}{1 - q^{2m}} - \sum_{m = 1}^{\infty}\frac{q^{2m}e^{-2imz}}{1 - q^{2m}}\right)\notag\\ &= \cot z + 4\sum_{n = 1}^{\infty}\frac{q^{2n}}{1 - q^{2n}}\sin 2nz\tag{3}\end{align}

---

The above proof is taken from one of my blog posts (see equation $(25)$ of that post).