showing that this theta series is holomorphic

Question

Consider the theta series $$f(w)=\sum_{n \in \mathbb{Z}}e^{\pi i(n+w)^2z}$$for fixed $z\in \mathbb{H}$. Then f(w) is holomorphic in $\mathbb{C}.$
Let $z=x+iy,y>0,w=u+iv$ then $$|e^{\pi i(n+w)^2z}|=e^{-(n^2y+2n(uy+vx)+u^2y+2uvx-v^2y)}. $$ Can I use then $ n^2y+2n(uy+vx)+u^2y+2uvx-v^2y>n^2y/2$ for $n\geq n_0$, so that I can estimate this series by the geometric series to finally use the weierstrass m-test?

Answer 1

Yes, you can use your estimate. However, since the sum is over all $n \in \mathbb{Z}$ you need it for $\lvert n\rvert \geqslant n_0$, not only for $n \geqslant n_0$. The value $n_0$ cannot be chosen independently of $w$, but for every $R > 0$ it can be chosen such that the inequality holds uniformly for all $w$ with $\lvert w\rvert \leqslant R$. And we can also let $z$ vary in the upper half-plane, for every compact $K \subset \mathbb{H}$ and every $R > 0$ there is an $n_0$ and a $\delta > 0$ such that $$\bigl\lvert e^{\pi i(n+w)^2z}\bigr\rvert \leqslant e^{-n^2\delta}$$ holds for all $\lvert n\rvert \geqslant n_0$ and all $z \in K$, $\lvert w\rvert \leqslant R$.

Thus by the Weierstraß $M$-test the series converges absolutely and locally uniformly on $\mathbb{C}\times \mathbb{H}$, and $$g(w,z) = \sum_{n \in \mathbb{Z}} e^{\pi i(n+w)^2z}$$ is holomorphic on $\mathbb{C}\times \mathbb{H}$.