Let $\vartheta(z)=\sum_{n\in Z}e^{\pi \mathrm{i}n^2z}$ I want to show that $$(\vartheta\left(\frac{z}{4}\right)-\vartheta(z))=\sum_{n \in \mathbb{Z}}e^{\pi\mathrm{i}(n+1/2)^2z}$$ After using the hint.
\begin{align*} \vartheta\left(1-\frac{1}{z}\right)&=\sqrt{\frac{z}{i}}\left(\vartheta\left(\frac{z}{4}\right)-\vartheta(z)\right)=\sqrt{\frac{z}{i}}\sum_{n \in \mathbb{Z}}e^{\pi\mathrm{i}(2n)^2(z/4)}+e^{\pi \mathrm{i}(2n+1)^2(z/4)}-e^{\pi \mathrm{i}n^2z}\\&=\sqrt{\frac{z}{i}}\sum_{n \in \mathbb{Z}}e^{\pi \mathrm{i}n^2z+\pi \mathrm{i}nz+\pi \mathrm{i}(z/4)}=\sqrt{\frac{z}{i}}\sum_{n \in \mathbb{Z}}e^{\pi\mathrm{i}(n+1/2)^2z}. \end{align*}
HINT:
Write $$\Theta(z/4)=\sum_{n\in \mathbb{Z}}\left( e^{i\pi (2n)^2(z/4)}+e^{i(2n+1)^2(z/4)}\right)$$in terms of even and odd terms.