$ \sum_{n=1}^\infty \csc^2(\omega\pi n)= \frac{A}{\pi} +B $ $$ \sum_{n=1}^\infty \csc^2(\omega\pi n)= \frac{A}{\pi} +B $$ if $\omega =-\frac{1}{2}+\frac{\sqrt{3}}{2}i$ find $\frac{A^2}{B^2}$
My Attempt
$$ \sum_{n=1}^\infty \csc^2(\omega\pi n)= \sum_{n=1}^\infty csch^2(i\omega\pi n)= 4\sum_{n=1}^\infty  \big(e^{\pi n \big( \frac{i}{2} + \frac{ \sqrt{3} }{2} \big) }-e^{-\pi n \big( \frac{i}{2} + \frac{ \sqrt{3} }{2} \big)}\big) ^{-2} $$ $$\sum_{n=1}^\infty  \big(e^{\pi n \big( \frac{i}{2} + \frac{ \sqrt{3} }{2} \big) }-e^{-\pi n \big( \frac{i}{2} + \frac{ \sqrt{3} }{2} \big)}\big) ^{-2}= \big(ie^{\pi\frac{\sqrt{3}}{2}}+ie^{-\pi\frac{\sqrt{3}}{2}}\big)^{-2}+ \big(-e^{\pi\sqrt{3}}+e^{-\pi\sqrt{3}}\big)^{-2} +\big(-ie^{3\pi\frac{\sqrt{3}}{2}}-ie^{-3\pi\frac{\sqrt{3}}{2}}\big)^{-2}+  \big(e^{2\pi\sqrt{3}}-e^{-2\pi\sqrt{3}}\big)^{-2} +...$$$$=  \sum_{n=0}^\infty  \big(ie^{(4n+1)\pi\frac{\sqrt{3}}{2}}+ie^{-(4n+1)\pi\frac{\sqrt{3}}{2}}\big)^{-2} +\sum_{n=0}^\infty  \big(-e^{(2n+1)π√3}+e^{-(2n+1)π√3}\big)^{-2} +\sum_{n=0}^\infty  \big(-ie^{(3+4n)\pi\frac{\sqrt{3}}{2}}+-ie^{-(4n+3)\pi\frac{\sqrt{3}}{2}}\big)^{-2}+
\sum_{n=0}^\infty  \big(e^{(2n)π√3}-e^{-(2n)π√3}\big)^{-2}  $$ $$\sum_{n=0}^\infty  \big(-e^{(4n+1)\pi\sqrt{3}}-2-e^{-(4n+1)\pi\sqrt{3}}\big)^{-1}+ \sum_{n=0}^\infty   \big(e^{2(2n+1)\pi\sqrt{3}}-2+e^{-2(2n+1)\pi\sqrt{3}}\big)^{-1} + \sum_{n=0}^\infty  \big(e^{(3+4n)\pi\sqrt{3}}-2+e^{-(3+4n)\pi\sqrt{3}}\big)^{-1}+ \sum_{n=1}^\infty  \big(e^{4n\pi\sqrt{3}}-2+e^{-4n\pi\sqrt{3}}\big)^{-1}  $$ I have found the sums numerically and $\sum_{n=0}^\infty  \big(-e^{(4n+1)\pi\sqrt{3}}-2-e^{-(4n+1)\pi\sqrt{3}}\big)^{-1}+ \sum_{n=0}^\infty   \big(e^{2(2n+1)\pi\sqrt{3}}-2+e^{-2(2n+1)\pi\sqrt{3}}\big)^{-1} + \sum_{n=0}^\infty  \big(e^{(3+4n)\pi\sqrt{3}}-2+e^{-(3+4n)\pi\sqrt{3}}\big)^{-1}+ \sum_{n=1}^\infty  \big(e^{4n\pi\sqrt{3}}-2+e^{-4n\pi\sqrt{3}}\big)^{-1} \approx -0.00429$
How can I evaluate this analytically?

Here is an approach using the residue theorem.
The residues of $\newcommand{\res}{\operatorname*{Res}}f(z)=\pi\cot\pi z\csc^2\omega\pi z$ at its poles are:
$$\res_{z=0}f(z)=\frac{1-\omega}{3};\quad\res_{z=n}f(z)=\csc^2\omega\pi n;\quad\res_{z=n/\omega}f(z)=-\omega\csc^2\omega\pi n\quad(n\in\mathbb{Z}_{\neq 0})$$ (easy to obtain using power series; we keep in mind $\omega^3=1\implies 1/\omega=-1-\omega$).
Now consider $\int_{\Gamma}f(z)\,dz$ where $\Gamma$ is the boundary of the parallelogram
$\hspace{3cm}$
$$\big\{z : \max\{|\Re z|,|\Re(\omega z)|\}\leqslant N+1/2\big\}$$
and take $N\to\infty$. On $AB$ or $CD$ we have $z=\pm\frac1\omega\left(N+\frac12+it\right)$, with "$+$" on $CD$ and "$-$" on $AB$, and $t\in\mathbb{R}$; using $\lim\limits_{y\to-\infty}\cot(x+iy)=i$ (uniformly w.r.t. $x\in\mathbb{R}$), we see that both $\int_{AB}$ and $\int_{CD}$ tend to $$-\frac\pi\omega\int_{-\infty}^\infty\frac{dt}{\cosh^2\pi t}=-\frac2\omega$$ as $N\to\infty$. The integrals along $BC$ and $DA$ both tend to $0$, hence the residue theorem gives $$\frac{1-\omega}{3}+2(1-\omega)\sum_{n=1}^\infty\csc^2\omega\pi n=-\frac{2}{\omega\pi i}\implies\color{blue}{\sum_{n=1}^\infty\csc^2\omega\pi n=\frac{1}{\pi\sqrt3}-\frac16}.$$
