TITLE: Theta function identity
QUESTION [5 upvotes]: Let us consider the Theta function
$$\theta(\tau) =  \sum_{n \in \mathbb{Z}} e^{\pi i n^2 \tau} \text{ for } \mathrm{Im}(\tau)>0.$$
Then it is rather easy to see that 
$$\theta(\tau+2)=\theta(\tau)$$
and
$$\theta(\tau)+\theta(\tau+1)=2 \cdot \theta(4 \tau).$$
Further, one can prove that
$$\theta(-1/ \tau) = \sqrt{\frac{\tau}{i}} \cdot \theta(\tau).$$
All together, this yields
$$\theta(1- \frac{1}{\tau}) = \theta(- \frac{1}{\tau}) + \theta(1- \frac{1}{\tau}) - \theta(- \frac{1}{\tau}) = 2 \cdot \theta(- \frac{4}{\tau}) - \theta(- \frac{1}{\tau}) = \sqrt{\frac{\tau}{i}} \cdot (\theta(\frac{\tau}{4}) - \theta(\tau)).$$
Here $\sqrt{}$ always denotes the principal branch of the square root. 
Next, my textbook claims that 
$$f(\tau) := \theta^4(\tau)-\theta^4(\tau+1)+\tau^{-2} \cdot \theta^4(1- \frac{1}{\tau})$$
fulfils both $f(\tau+1) = - f(\tau)$ and $f(- \frac{1}{\tau}) = - \tau^{2}  \cdot f(\tau)$ (and uses this to deduce that actually $f = 0$). It should be possible to derive these using the above $\theta$-identities, but I am completely stuck here. Any help is appreciated!

REPLY [4 votes]: Let us compute explicitly:
$$
\begin{aligned}
f(\tau)+f(\tau+1)
&=
\theta^4(\tau)-\theta^4(\tau+1)
+\tau^{-2} \cdot \theta^4\left(1- \frac{1}{\tau}\right)
\\
&\qquad\qquad
+
\theta^4(\tau+1)-\theta^4(\tau+2)
+(\tau+1)^{-2} \cdot \theta^4\left(1- \frac{1}{\tau+1}\right)
\\
&=
\tau^{-2} \cdot \theta^4\left(1- \frac{1}{\tau}\right)
+(\tau+1)^{-2} \cdot \theta^4\left(1- \frac{1}{\tau+1}\right)
\\
&=
\tau^{-2} \cdot\left[\  \theta^4\left(\frac{\tau-1}{\tau}\right)
+\left(\frac\tau{\tau+1}\right)^2 \cdot \theta^4\left(\frac{\tau}{\tau+1}\right)
\ \right]
\\
&=
\tau^{-2} \cdot\left[\  \theta^4\left(\frac{\tau-1}{\tau}-2\right)
-\theta^4\left(-\frac 1{\qquad\frac{\tau}{\tau+1}\qquad}\right)
\ \right]
\\
&=0\ ,
\\[3mm]
f\left(-\frac1\tau\right)
&=
\theta^4\left(-\frac1\tau\right)
-
\theta^4\left(-\frac1\tau+1\right)
+
\left(-\frac1\tau\right)^{-2}\theta^4\left(1-\frac1{-1/\tau}\right)
\\
&=
\theta^4\left(-\frac1\tau\right)
-
\theta^4\left(1-\frac1\tau\right)
+
\tau^2\theta^4(\tau+1)
\\
&=
-\tau^2\theta^4(\tau)
+\tau^2\theta^4(\tau+1)
-
\tau^2\cdot\tau^{-2}
\theta^4\left(1-\frac1\tau\right)
\\
&=
-\tau^2\;f(\tau)\ .
\end{aligned}
$$
(The power of $\tau$ differs in the last relation.)