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For the $\theta$-function $$\theta (z) = \sum_{n \in \mathbb{z}} q^{n^2}e^{2\pi inz},$$ for $q$ given by $e^{\pi i\tau}$ for some $\tau \in \mathbb{C}$ with $Im(\tau) > 0$, suppose we've proved the results $\theta(z+1) = \theta (z)$ and $\theta(z+\tau)=q^{-1}e^{-2\pi iz}\theta(z)$ for any $z \in \mathbb{C}$. Is it possible to prove that $$\theta(z+\tau^*) = -e^{-2\pi iz}\theta(\tau^* - z)$$ for any $z \in \mathbb{C}$ where $\tau^* = \frac{1+\tau}{2}$? I'm assuming you can use a trick where you change $n$ to $n-1$ or $-n$ in one of the sums to get the other, but I haven't been able to puzzle out the details yet and there's probably a quick proof out there.

I mentioned the earlier two identities because they're clearly similar, but it's hard to see how to apply them when we have a fraction in the argument and not an ordinary linear combination of $1$ and $\tau$. Any help would be appreciated.

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If you write out the expansion of $\theta(z+\tau^*)$ you have

$$ \theta(z+\tau^*) = \sum_{n=-\infty}^{+\infty}(-1)^n q^{n(n+1)} e^{2\pi i n z}$$

If you write out $-e^{-2\pi i z} \theta(\tau^* -z)$ you have

$$-e^{-2\pi i z} \theta(\tau^* -z) = -e^{-2\pi i z} \sum_{n=-\infty}^{+\infty} (-1)^n q^{n^2 + n} e^{-2\pi i n z}$$

If take this and change the summation variable $n\rightarrow -n$, you have

$$ -e^{-2\pi i z} \theta(\tau^* -z) = -e^{-2\pi i z} \sum_{n=-\infty}^{+\infty} (-1)^n q^{n^2 -n} e^{2\pi i n z} = \sum_{n=-\infty}^{+\infty} (-1)^{n-1} q^{n^2 -n } e^{2\pi i z (n-1)}$$

Now if we make the change of variable $n\rightarrow m+1$ we have

$$\sum_{n=-\infty}^{+\infty} (-1)^{n-1} q^{n^2 -n } e^{2\pi i z (n-1)} = \sum_{m=-\infty}^{+\infty} (-1)^m q^{m(m+1)}e^{2\pi i m z}$$

and so the two expressions are equal.

I hope this answers your question.

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