I want to compute the residue integral for the $q$-theta function, and derive its properties. First, I'll briefly explain the definition
\begin{align} & \theta(a;q)=(a;q)(q/a;q)=\prod_{i=0}^{\infty}(1-aq^i)(1-a^{-1}q^{i+1}), \qquad (q;q) =\prod_{i=0}^{\infty}(1-q^{i+1}) \end{align} where $(a;q)=\prod_{i=0}^{\infty}(1-aq^i).$
Now I'd like to check the following properties:
$\begin{align} \theta(qz; q)&=-\frac{1}{z}\theta(z;q),\\ \theta(z^{-1};q)&=\theta(qz;q)=-\frac{1}{z}\theta(z;q), \end{align}$
and I am not sure about the last property which follows:
$\theta'(1;q)=-(q;q)^2.$
Is it true?
I tried to manipulate the equation from the definition, but even the first property does not seem easy... If you know about this function, please give me hint or explicit calculation.
The function you are studying comes up a lot in $q$-series. The first property of $(z;q)$, the $q$-Pochhammer symbol, is $(zq;q) = 1/(1-z)(z;q)$ and also $(q/z;q)=z/(z-1)(1/z;q)$. Using this, it is easy to deduce the first property of $q$-theta. First, using the definition of $q$-theta gives $$\theta(qz;q) = (qz;q)(1/z;q) = 1/(1-z)(z;q)(z-1)/z(q/z;q) = -\theta(z;q)/z.$$ Second, using the definition of $q$-theta again gives $$\theta(1/z;q)=(1/z;q)(zq;q)=-(1-z)/z(q/z;q)(zq;q)=-(q/z;q)(z;q)/z=-\theta(z;q)/z.$$ The third property is true, and one way to prove it is to express $q$-theta in terms of Ramanujan theta $$\theta(z;q)(q;q)=f(-z,-q/z)= -(z-1)+(z^2-1/z)q-(z^3-1/z^2)q^3+(z^4-1/z^3)q^6+\dots$$ and thus $d/dz$ and evaluating at $z=1$ both sides gives $$\theta'(1;q)(q;q)=-1+3q-5q^3+7q^6-9q^{10}+\dots=-(q;q)^3$$ where the right right is a well known result and thus proving $\theta'(1;q)=-(q;q)^2$.