Faddeev’s quantum dilogarithm $\Phi_b(z)$ is defined by the integral
$$\Phi_b(z):=\exp\left(\int_{i0-\infty}^{i0+\infty}\frac{e^{-i2zw} }{4\sinh(wb)\sinh\left(\frac{w}{b}\right)w}dw \right) \tag{1} $$
in the strip $|\Im(z)| <|\Im(c_b)| $ where $$c_b := \frac{i(b+b^{-1})}{2}. $$
It seems well-known (for example, see the Appendix A of A TQFT from Quantum Teichmüller Theory by Jørgen Ellegaard Andersen and Rinat Kashaev) that when $\Im(b^2)>0$, the integral can be calculated explicitly as
$$ \Phi_b(z)= \frac{(e^{2\pi(z+c_b)b};q^2)_\infty}{(e^{2\pi(z-c_b)b^{-1}};\bar{q}^2)_\infty}, \tag{2}$$
where $q = e^{2\pi i b^2}$, $\bar{q}= e^{-2\pi ib^{-2}}$ and $(x;q)_\infty$ is the $q$-Pochhammer symbol
$$(x;q)_\infty := \prod_{k=0}^\infty (1-q^kx). $$
With definition (1), clearly we have $\Phi_{b}=\Phi_{-b}$. However, with formula (2), it is easy to deduce that $\Phi_b$ is meromorphic with
$$ \text{poles: }c_b+i\mathbb{N}b+i\mathbb{N}b^{-1},\quad \text{zeros: }-c_b-i\mathbb{N}b -i\mathbb{N}b^{-1}, \tag{3}$$
where $\mathbb{N}=\mathbb{Z}_{\geq 0}$.
With the only restriction being $\Im(b^2)>0$, which means that $b$ is either in the first quadrant or in the third quadrant, it is possible that the formula (2) is valid for both $b$ and $-b$, so that, substituting $-b$ into (3), $\Phi_{-b}$ would be meromorphic with
$$ \text{poles: }-c_b-i\mathbb{N}b-i\mathbb{N}b^{-1},\quad \text{zeros: }c_b+i\mathbb{N}b +i\mathbb{N}b^{-1}.$$
Therefore $\Phi_b=\Phi_{-b}$ (by (1)) while their poles and zeros are different (by (2))! How could this be possible? Is it me or is there some condition that is missing?
Thanks in advance for any help.
My advisor just told me that $\Im (b)\geq 0$ is always assumed when considering Faddeev’s quantum dilogarithm.
So formula (2) is valid only under the condition that $\Im (b)\geq 0$ and $\Im (b^2)>0$. It seems that in the appendix I mentioned above the order of introducing the conventional $\Im(b)\geq 0$ and formula (2) is reversed, leading to this unfortunate confusion.