$q$-digamma function evaluation What is the value of $\psi_2^{(0)}(1)$, where $\psi_q^{(0)}(z)$ is the $q$-digamma function?
My attempt:
\begin{align*}
\psi_2^{(0)}(z) &=\frac{1}{\Gamma_2(z)}\frac{d\Gamma_2(z)}{dz}
\\&=\frac{(2^z\ ;2)_\infty}{(2\ ;2)_\infty(1-2)^{1-z}}\cdot\frac{d}{dz}\left((1-2)^{1-z}\frac{(2\ ;2)_\infty}{(2^z\ ;2)_\infty}\right)
\\
&=\frac{1}{(-1)^{1-z}}\prod_{n=0}^\infty\frac{1-2^{n+z}}{1-2^{n+1}}\cdot\frac{d}{dz}(-1)^{1-z}\prod_{n=1}^\infty\frac{1-2^{n+1}}{1-2^{n+z}},
\end{align*}
where $\Gamma_q(z)$ is the q-gamma function, and $(a\ ;q)_k$ is the q-Pochhammer symbol. Evaluating at $1$, we see
$$\begin{align*}
\psi_2^{(0)}(1)&=\frac{1}{(-1)^0}\prod_{n=0}^\infty\frac{1-2^{n+1}}{1-2^{n+1}}\cdot\frac{d}{dz}(-1)^{1-z}\prod_{n=1}^\infty\frac{1-2^{n+1}}{1-2^{n+z}}\Bigr|_{z=1}
\\&=\frac{d}{dz}(-1)^{1-z}\prod_{n=1}^\infty\frac{1-2^{n+1}}{1-2^{n+z}}\Bigr|_{z=1}
\\&=\frac{d}{dz}\exp\left((\ln(-1)^{1-z}+\sum_{n=0}^\infty\ln\left(\frac{1-2^{n+1}}{1-2^{n+z}}\right)\right)\Bigr|_{z=1}
\\&=\exp\left(\ln(-1)^{1-z}+\sum_{n=0}^\infty\ln\left(\frac{1-2^{n+1}}{1-2^{n+z}}\right)\right)\left(i\pi+\sum_{n=0}^\infty\frac{2^{n+z}\ln2}{1-2^{n+z}}\right)\Bigr|_{z=1}
\\&=i\pi+\ln2\sum_{n=0}^\infty\frac{2^{n+1}}{1-2^{n+1}}
\end{align*}$$
which clearly diverges.... and even if it converged, I would have an imaginary part. Wolfram Alpha gives only an approximation of about $-0.7671026$, so what am I missing here?

The q-Gamma function there is  another definition for q>1. The q-gamma function is defined for $x>0$ and $0<q<1$ by $\Gamma_q(x)=(1-q)^{1-x}\prod_{n\geq 0}{\frac{1-q^{n+1}}{1-q^{n+x}}}$. For $q>1$ the q-gamma function is defined $\Gamma_q(x)=q^{\frac{(x-1)(x-2)}{2}}\Gamma_{1/q}(x).$
