Let's use the classical definition $$ \vartheta _1\left( z,q \right) =-i\sum_{n\in \mathbb{Z}}{\left( -1 \right) ^nq^{\left( n+\frac{1}{2} \right) ^2}e^{i\left( 2n+1 \right) z}}\,\,\,\,\,\ q=e^{i\pi \tau} $$ Now consider function \begin{equation} g\left( z \right) =\frac{\vartheta _1\left( z-a,q^{1/3} \right)}{\vartheta _{1}^{2}\left( z,q \right) \vartheta _1\left( z-3a,q \right)}\,\,\,\,\,0<a<\frac{\pi}{3} \end{equation} It is easy to verify that it is an elliptic function with periods $\pi $ and $\pi \tau $.In a periodic parallelogramand ,the poles of $g\left( z \right) $ are $z=0$ and $z=3a$， and the residue theorem for elliptic functions tell us $$ \mathrm{res}\left( g,0 \right) +\mathrm{res}\left( g,3a \right) =0 $$ it’s easy to calculate $$ \mathrm{res}\left( g,3a \right) =\frac{\vartheta _1\left( 2a,q^{1/3} \right)}{\vartheta _{1}^{2}\left( 3a,q \right) \vartheta _{1}^{\prime}\left( q \right)} $$ then consinder function $$ G\left( z \right) =z^2g\left( z \right) $$ and we can know $ \mathrm{res}\left( g,0 \right) =G^{\prime}\left( 0 \right) $ ,take the logarithmic derivative of both sides of this function we have $$ \frac{G^{\prime}}{G}\left( z \right) =\frac{2}{z}-2\frac{\vartheta _{1}^{\prime}}{\vartheta _1}\left( z,q \right) +\frac{\vartheta _{1}^{\prime}}{\vartheta _1}\left( z-a,q^{1/3} \right) -\frac{\vartheta _{1}^{\prime}}{\vartheta _1}\left( z-3a,q \right) $$ clearly,$$ G\left( 0 \right) =\frac{\vartheta _1\left( a,q^{1/3} \right)}{\vartheta _{1}^{\prime}\left( q \right) ^2\vartheta _1\left( 3a,q \right)} $$ so combined with the above results，we have $$ \frac{\vartheta _1\left( 2z,q^{1/3} \right) \vartheta _{1}^{\prime}\left( q \right)}{\vartheta _1\left( 3z,q \right) \vartheta _1\left( z,q^{1/3} \right)}=\frac{\vartheta _{1}^{\prime}}{\vartheta _1}\left( z,q^{1/3} \right) -\frac{\vartheta _{1}^{\prime}}{\vartheta _1}\left( 3z,q \right) $$ Using Fourier series $$ \frac{\vartheta _{1}^{\prime}}{\vartheta _1}\left( z,q \right) =\cot \left( z \right) +4\sum_{n=1}^{\infty}{\frac{q^{2n}}{1-q^{2n}}\sin \left( 2nz \right)} $$ we have $$ \frac{\vartheta _1\left( 2z,q^{1/3} \right) \vartheta _{1}^{\prime}\left( q \right)}{\vartheta _1\left( 3z,q \right) \vartheta _1\left( z,q^{1/3} \right)}=\cot \left( z \right) -\cot \left( 3z \right) +4\sum_{n=1}^{\infty}{\frac{q^{\frac{2n}{3}}}{1-q^{\frac{2n}{3}}}\sin \left( 2nz \right)}-4\sum_{n=1}^{\infty}{\frac{q^{2n}}{1-q^{2n}}\sin \left( 6nz \right)} $$ This formula can give some interesting results ,for example just use transform $ z=z+\frac{\pi}{2} $ we have $$ \vartheta _3\vartheta _4\vartheta _3\left( q^3 \right) \vartheta _4\left( q^3 \right) =1+4\sum_{n=1}^{\infty}{\frac{\left( -1 \right) ^nnq^{2n}}{1-q^{2n}}}-12\sum_{n=1}^{\infty}{\frac{\left( -1 \right) ^nnq^{6n}}{1-q^{6n}}} $$ in other words，we have $$ \sum_{m_i\in \mathbb{Z}}{^{\prime}\frac{\left( -1 \right) ^{m_3+m_4}}{\left( m_{1}^{2}+3m_{2}^{2}+m_{3}^{2}+3m_{4}^{2} \right) ^2}}=-\frac{\pi ^2}{9}\log \left( 2 \right) $$ Even though I got the right result, I couldn't get the right result using the same method for other functions. for example,consider $$ f\left( z \right) =\frac{\vartheta _1\left( 11z-a,q^{11} \right)}{\vartheta _{1}^{2}\left( z,q^{1/5} \right) \vartheta _1\left( z-a,q \right)}\,\,\left( 0<a<\pi \right) $$ the poles of $f\left( z \right) $ are $z=0$ and $z=a$,use the same method we have $$ \frac{\vartheta _1\left( 10a,q^{11} \right) \vartheta _1\left( a,q \right) \vartheta _{1}^{\prime}\left( q^{1/5} \right) ^2}{\vartheta _{1}^{2}\left( a,q^{1/5} \right) \vartheta _1\left( a,q^{11} \right) \vartheta _{1}^{\prime}\left( q \right)}=11\frac{\vartheta _{1}^{\prime}}{\vartheta _1}\left( a,q^{11} \right) -\frac{\vartheta _{1}^{\prime}}{\vartheta _1}\left( a,q \right) $$
but this result is wrong.
My question:why can't I get the right result，whether I left out other poles? anyone can help me？