Vtyjkyuk

For the $theta$-function $$theta (z) = sum_n in mathbbz q^n^2e^2pi inz,$$
for $q$ given by $e^pi itau$ for some $tau in mathbbC$ with $Im(tau) > 0$, suppose we've proved the results $theta(z+1) = theta (z)$ and $theta(z+tau)=q^-1e^-2pi iztheta(z)$ for any $z in mathbbC$. Is it possible to prove that $$theta(z+tau^*) = -e^-2pi iztheta(tau^* - z)$$ for any $z in mathbbC$ where $tau^* = frac1+tau2$? 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.

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

$$ theta(z+tau^*) = sum_n=-infty^+infty(-1)^n q^n(n+1) e^2pi i n z$$

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

$$-e^-2pi i z theta(tau^* -z) = -e^-2pi i z sum_n=-infty^+infty (-1)^n q^n^2 + n e^-2pi i n z$$

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

$$ -e^-2pi i z theta(tau^* -z) = -e^-2pi i z sum_n=-infty^+infty (-1)^n q^n^2 -n e^2pi i n z = sum_n=-infty^+infty (-1)^n-1 q^n^2 -n e^2pi i z (n-1)$$

Now if we make the change of variable $nrightarrow m+1$ we have

$$sum_n=-infty^+infty (-1)^n-1 q^n^2 -n e^2pi i z (n-1) = sum_m=-infty^+infty (-1)^m q^m(m+1)e^2pi i m z$$

and so the two expressions are equal.

I hope this answers your question.