Show that $f(qz) = qz^2 f(z)$ where $f(z) = [\theta(z) \theta(1/z)]^{-1}$

Question

Let $\theta(z) = (z;q)(q/z; q)$ where $(q;z) = \prod_{i=0}^\infty (1 - zq^i) $. Then let $f(z)$ be defined by:

$$ f(z) = \frac{1}{\theta(z) \theta(1/z)} $$

Show that $\boxed{\color{#0033FF}{f(qz) = qz^2 f(z)}}$ It's hard to believe I am thrown off by this algebra, but here is my work:

$$ \frac{f(qz)}{f(z)} = \frac{\theta(zq)\theta(\frac{1}{qz})}{\theta(z)\theta(1/z)} = \frac{(qz;q)(\frac{1}{z};q)(\frac{1}{qz};q)(q^2 z;q)}{(z;q)(\frac{q}{z};q)(\frac{1}{z};q)(qz;q)} = \frac{(\frac{1}{qz};q)(q^2 z;q)}{(z;q)(\frac{q}{z};q)}$$

Sorry if that was harsh on the eyes. This doesn't look at all like it should be $q^2 z$. Then I started making some identities:

$$ (qz;q) = \prod_{i=0}^\infty (1 - zq^{i+1}) = \frac{\prod_{i=0}^\infty (1 - zq^i)}{1 - zq} = \frac{(z;q)}{1 - zq}$$

Then I sort of gave up. I don't see how the cancellation can occur. Any ideas? Is the relation false?

Answer 1

$$ (qz;q) = \prod_{i=0}^\infty (1 - zq^{i+1}) = \frac{\prod_{i=0}^\infty (1 - zq^i)}{1 - z} = \frac{(z;q)}{1 - z}$$ $$ (q^2z;q) = \prod_{i=0}^\infty (1 - zq^{i+2}) = \frac{\prod_{i=0}^\infty (1 - zq^i)}{(1-zq)(1 - z)} = \frac{(z;q)}{(1-zq)(1 - z)}$$

$$ (\frac1{qz};q) = \prod_{i=0}^\infty (1 - \tfrac1zq^{i-1}) = ({1 - \tfrac1{zq}})\prod_{i=0}^\infty (1 - \tfrac1zq^i) = {(\tfrac1z;q)}({1 - \tfrac1{zq}})$$

$$ \frac{f(qz)}{f(z)} = \frac{\theta(z)\theta(\tfrac{1}{z})}{\theta(zq)\theta(\tfrac{1}{qz})} = \frac{(z;q)(\tfrac{q}{z};q)(\tfrac{1}{z};q)(qz;q)}{(qz;q)(\tfrac{1}{z};q)(\tfrac{1}{qz};q)(q^2 z;q)} = \frac{(z;q)(\tfrac{q}{z};q)}{(\tfrac{1}{qz};q)(q^2 z;q)} = \frac{(\tfrac{q}{z};q)(1-zq)(1 - z)}{(\tfrac{1}{qz};q)} = \frac{(\tfrac{1}{z};q)(1-zq)(1 - z)}{(\tfrac{1}{qz};q)(1-\frac1z)} = \frac{(1-zq)(1 - z)}{(1-\frac1z)({1 - \tfrac1{zq}})} = \frac{(1-zq)(1 - z)}{\frac1z(z-1)\tfrac1{zq}({zq - 1})}={z^2q} $$