In arXiv:math/0309252v4 at the bottom of page 11, the following result is proposed
$$ b_1 \theta(b_0 b_1^{\pm};p) \frac{z^{-1}\theta(z^2;p)}{\theta(b_0 z^{\pm};p)\theta(b_1 z^{\pm};p)} = (p;p)^{-2}\sum_{k\neq 0}\frac{b_0^k + (p/b_0)^k - b_1^k - (p/b_1)^k}{1-p^k} z^k $$
Where
$$ \theta(z;p)=\prod_{k=0}^{\infty} (1-p^k z)(1-p^{k+1}/z), \quad \theta(bz^{\pm};p)=\theta(bz;p)\theta(b/z;p) $$ To prove this, the author writes the integral of the RHS of the above as the following function
$$ \log{\frac{\theta(b_0z^{\pm};p)}{\theta(b_1z^{\pm};p)}}$$
and then by comparing poles, he states that the following must be true
$$ z \frac{d}{dz}\log{\frac{\theta(b_0z^{\pm};p)}{\theta(b_1z^{\pm};p)}} = C(b_0,b_1,p) \frac{z^{-1}\theta(z^2;p)}{\theta(b_0z^{\pm};p)\theta(b_1z^{\pm};p)}$$
Where the coefficient $C(b_0,b_1,p)$ can be computed by comparing asymptotics.
I am struggling with this last line, where comparing poles allows one to immediately say that this experssion on the LHS has to be equal to this nontrivial combination of theta functions on the RHS. Surely it could be some other combination of theta functions? What constrains the numerator to be $z^{-1}\theta(z^2;p)$ (is it not possible to get something like $\theta(p^{1/2}z^2;p)$, or would that correspond to a different RHS?) ?
