Ellptic\Jacobi theta function and its residue integral The Ellptic\Jacobi theta function is given by
\begin{align}
  \theta_1(\tau|z)&=\theta_1(q,y)=-iq^{\frac{1}{8}}y^{\frac{1}{2}}\prod_{k=1}^{\infty}(1-q^k)(1-yq^k)(1-y^{-1}q^{k-1}) \\
  &= -i\sum_{n\in \mathbb{Z}}(-1)^n e^{2\pi i z(n+\frac{1}{2})} e^{\pi i \tau(n+\frac{1}{2})^2}
\end{align}
Where $q=e^{2\pi i \tau}$, and $y=e^{2\pi i z}$ 
and dedkind eta function is given as
$\eta(\tau)=\eta(q)=q^{\frac{1}{24}} \prod_{n=1}^{\infty}(1-q^n) $

Now i want to evaluate following  relation
$\theta'_1(\tau|0)=\theta'_1(q,1)=2\pi \eta(q)^3$
The results is well known (i guess...) but i can not find any papers or textbooks contain some procedure of this. 
 A: Rewrite the product in your first equation as $-iq^{1/8}y^{1/2}(1-y^{-1})\prod_{k=1}^{\infty}(1-q^k)(1-yq^k)(1-y^{-1}q^k)$ (just tweak the third part of the product), divide out $(1-y^{-1})$ from the equation, then use L'Hôpital's rule (with respect to $z$ as it approaches $0$) to calculate the limit of the left-hand theta expression; the result you want follows immediately.
You can also substitute $y=e^{2i\pi/3}$ and $q^{1/3}\mapsto q$ to evaluate the Dedekind eta function as a theta function (since $(1-q^{k/3}e^{2i\pi/3})(1-q^{k/3}e^{-2i\pi/3})(1-q^{k/3})=1-q^k$).
As far as I know, I don't think this problem needs residue calculus to solve it, so the title may be misleading.
A: From Antonio DJC's comment i explicit compute  
\begin{align}
&\theta_1(q, y)= -i y^{\frac{1}{2}}q^{\frac{1}{8}} \prod_{n=1}^\infty \left(1-q^n\right)\left(1-y q^n\right) \left(1-y^{-1}q^{n-1} \right) \\
&\phantom{\Theta(q,y)}= -i y^{\frac{1}{2}}q^{\frac{1}{8}}(1-y^{-1}) \prod_{n=1}^\infty \left(1-q^n\right)\left(1-y q^n\right) \left(1-y^{-1}q^{n} \right),  \\
&\frac{\theta_1(q,y)}{1-y^{-1}}= -i y^{\frac{1}{2}} q^{\frac{1}{8}} \prod_{n=1}^\infty \left(1-q^n\right)\left(1-y q^n\right) \left(1-y^{-1}q^{n} \right) \\
& \lim_{z \rightarrow 0} \frac{\theta_1(\tau|z)}{1-y^{-1}} =\lim_{z \rightarrow 0} \frac{\theta_1'(\tau|z)}{y^{-2} 2\pi i} =  \frac{\theta_1'(q,1)}{2\pi i}
= -i q^{\frac{1}{8}} \prod_{n=1}^{\infty} (1-q^n)^3 = -i \eta(q)^3
\end{align}
where we used $\frac{df(y)}{dz} =  \frac{df}{dz} \frac{dy}{dz} = \frac{df}{dz} 2\pi i$.  And Dedkind eta function is given as
$\begin{align}
 \eta(q)=q^{\frac{1}{24}} \prod_{n=1}^\infty (1-q^n)
\end{align}$
