Riemann theta function - simple question about equation verification I basically have a simple algebra question. Olivier Debarre in his book on abelian varieties defines the Riemann theta function $\theta[a,b](z)$ for real numbers $a,b$ as in the picture below. 
Let $\tau \in \mathbb{C}$ be a complex, non-real, number, and $\Gamma_\tau=\mathbb{Z}\tau \oplus \mathbb{Z}$, so an element of $\Gamma_\tau$ is $p\tau + q$ for $p,q \in \mathbb{Z}$. Then Debarre claims that 
$$\theta[a,b](z+p\tau + q)=e^{\Box}\theta[a,b](z) $$
where you can see what goes in the box $\Box$ in the picture below. My question is why does this equation hold? I almost see why - do a change of variable in the index of summation $m$ by replacing $m$ by $m+p$ in the summation defining $\theta[a,b](z)$...this almost gives the result (the signs on 2pb and 2aq may need to be switched, I don't know) but I can't account for a $mq$ term.

 A: The trick, besides the $m \mapsto m+p$ change of variables, is to notice that $e^{2\pi i m q}=1$ since $m,q$ are integers. Here are the details:
$\theta(a,b)(z+ p\tau + q)$ is the sum over $m \in \mathbb{Z} $ of $e$ raised to $\pi i$ times
$$ \tau(m+a)^2 + 2(m+a)(z + \tau p + q +b) =$$
$$\tau(m+a)^2 + 2(m+a)\tau p + 2(m+a)(z + q +b) =$$
$$\tau[(m+a)^2 + 2(m+a) p] + 2(m+a)(z + q +b) =$$
$$\tau[(m+a)^2 + 2(m+a) p + p^2 -p^2] + 2(m+a )(z + q +b) =$$
$$\tau(m+p+a)^2 -\tau p^2 + 2(m+a )(z + q +b) =$$
$$\tau(m+p+a)^2 -\tau p^2 + 2[(m+a + p) - p][(z + b) +q] =$$
$$\tau(m+p+a)^2 -\tau p^2 + 2(m+a + p)(z + b)  +2(m+a+p)q - 2p(z+b) -2pq = $$
$$\tau(m+p+a)^2 -\tau p^2 + 2(m+a + p)(z + b)  +2mq + 2aq  - 2pz - 2pb $$
We can drop the $2mq$ term since it is an even integer that raises $e^{\pi i}$. So we are left with
$$\tau(m+p+a)^2 + 2(m+a + p)(z + b) -\tau p^2  + 2aq  - 2pz -2pb $$
i.e
$$\Large \theta(a,b)(z + p\tau +q)=\theta(a,b)(z) e^{\pi i(-\tau p^2  + 2aq  - 2pz -2pb) }$$
I'm getting a sign discrepancy from Debarre on the $2aq, 2pb$ terms, but close enough.
