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I'm studying $u(z,q):=\exp(\left[\tfrac{\pi q}{12}-\ln2-\sum_{k\ge1}\ln(1-e^{-2k\pi q})\right]z+\int_1^z\ln\varphi(x,q)\,dx)$ , with $$\varphi(z,q)=2e^{-\pi qz^2-\pi q/4}\sinh \pi qz\prod_{k\ge1}(1-e^{-2k\pi q})(1-e^{-2k\pi q+2\pi qz})(1-e^{-2k\pi q-2\pi qz})$$ With many symbolic manipulations I can show $\varphi(z+i/q,q)=-e^{\pi/q-2i\pi z}\varphi(z,q)$, and so \begin{align}u(z+i/q,q)&=\exp\left(\left[\tfrac{\pi q}{12}-\ln2-\sum_{k\ge1}\ln(1-e^{-2k\pi q})\right](z+i/q)+\int_1^{z+i/q}\ln\varphi(x,q)\,dx\right)\\ &=\exp\left(\left[\tfrac{\pi q}{12}-\ln2-\sum_{k\ge1}\ln(1-e^{-2k\pi q})\right](z+i/q)+\left[\int_1^z+\int_z^{z+i/q}\right]\ln\varphi(x,q)\,dx\right)\\ &=u(z,q)\exp\left(\left[\tfrac{\pi q}{12}-\ln2-\sum_{k\ge1}\ln(1-e^{-2k\pi q})\right](i/q)+\int_z^{z+i/q}\ln\varphi(x,q)\,dx\right) \end{align} But $\varphi(z,q)$ also has the form $$\frac2{\sqrt{q}}e^{-\pi/(4q)}\sin\pi z\prod_{k\ge1}(1-e^{-2k\pi/q})(1-e^{-2k\pi/q+2i\pi z})(1-e^{-2k\pi/q-2i\pi z})$$ due to the imaginary transformation $\varphi(z,q)=-\frac i{\sqrt{q}}e^{-\pi qz^2}\varphi(iqz,1/q)$ ($\varphi(z,q)$ is equivalent to $q^{-1/2}\vartheta_1(\pi z,e^{-\pi/q}$) and clearly it has antiperiod $1$, so similarly \begin{align}u(z+1,q)&=\exp\left(\left[\tfrac{\pi q}{12}-\ln2-\sum_{k\ge1}\ln(1-e^{-2k\pi q})\right](z+1)+\int_1^{z+1}\ln\varphi(x,q)\,dx\right)\\ &=\exp\left(\left[\tfrac{\pi q}{12}-\ln2-\sum_{k\ge1}\ln(1-e^{-2k\pi q})\right](z+1)+\left[\int_1^z+\int_z^{z+1}\right]\ln\varphi(x,q)\,dx\right)\\ &=u(z,q)\exp\left(\left[\tfrac{\pi q}{12}-\ln2-\sum_{k\ge1}\ln(1-e^{-2k\pi q})\right]+\int_z^{z+1}\ln\varphi(x,q)\,dx\right) \end{align} Now I consider the quotient $v(z,\alpha;q):=\frac{u(z+\alpha,q)}{u(z,q)}$ which, following the above calculations, equals $$\exp\left(\left[\frac{\pi q}{12}-\ln2-\sum_{k\ge1}\ln(1-e^{-2k\pi q})\right]\alpha+\int_z^{z+\alpha}\ln\varphi(x,q)\,dx\right)$$ Adding $i/q$ to $z$ leads to\begin{align}v(z+i/q,\alpha;q)&=\exp\left(\left[\frac{\pi q}{12}-\ln2-\sum_{k\ge1}\ln(1-e^{-2k\pi q})\right]\alpha+\int_{z+i/q}^{z+i/q+\alpha}\ln\varphi(x,q)\,dx\right)\\ &=\exp\left(\left[\frac{\pi q}{12}-\ln2-\sum_{k\ge1}\ln(1-e^{-2k\pi q})\right]\alpha+\int_z^{z+\alpha}\ln\left(-e^{\pi/q-2i\pi x}\varphi(x,q)\right)\,dx\right)\\ &=\exp\left(\left[\frac{\pi q}{12}-\ln2-\sum_{k\ge1}\ln(1-e^{-2k\pi q})\right]\alpha+\int_z^{z+\alpha}\left(\ln-1+\pi/q-2i\pi x+\ln\varphi(x,q)\right)\,dx\right)\\ &=\exp\left(\left[\frac{\pi q}{12}-\ln2-\sum_{k\ge1}\ln(1-e^{-2k\pi q})\right]\alpha+\alpha\ln-1+\alpha\pi/q-i\pi[(z+\alpha)^2-z^2]+\int_z^{z+\alpha}\ln\varphi(x,q)\,dx\right)\\ &=v(z,\alpha;q)(-1)^\alpha e^{\alpha\pi/q-\alpha^2i\pi-2\alpha i\pi z} \end{align} The last line above can then be divided by $\varphi^{\alpha}(z+i/q,q)=(-1)^{\alpha}e^{\alpha\pi/q-2i\pi\alpha z}\varphi^{\alpha}(z,q)$ to produce $$\frac{v(z,\alpha;q)}{\varphi^\alpha(z,q)}=e^{\alpha^2i\pi}\frac{v(z+i/q,\alpha;q)}{\varphi^\alpha(z+i/q,q)}$$ With similar calculations $$\frac{v(z,\alpha;q)}{\varphi^\alpha(z,q)}=\frac{v(z+1,\alpha;q)}{\varphi^\alpha(z+1,q)}$$ so if $\alpha^2$ is rational, I'll have a doubly periodic function in $z$ (and if necessary I can raise $\frac{v(z,\alpha;q)}{\varphi^\alpha(z,q)}$ to the power $k/\alpha$ to have simple-order poles).

My problem is I can't figure out any zeros or poles for $u(z,q)$ , thus I don't know if $v/\varphi^\alpha$ can properly be an elliptic function for $z$, even for $\alpha\in\mathbb{Z}$ (I do feel $u(z,q)$ must vanish somewhere.)

I'm most interested in the case $\alpha=1/4$ because, if I can evaluate $u(0,q)$ , then I can figure out $u(1/4,q)$ as $u(0,q)$ times some elliptic function at $z=0$.

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