I have the following integral:

$\displaystyle \displaystyle\int_{0}^{2\pi}\frac{1}{(4sinx+3i)^{2 }}= \left [{\begin{array}{ll}z=e^{ix} \,\,\\\frac{dz}{iz}=dx\end{array}\right]=$

since:

$\displaystyle \displaystyle sin x=\frac{1}{2i}(z-z^{-1})$

I rewrite the integral and get the following:

$\displaystyle \displaystyle=\int_{0}^{2\pi}\frac{1}{((\frac{2z^2-2}{iz})+3i)^{2}}\frac{dz}{iz}=$

$\displaystyle \displaystyle=\int_{0}^{2\pi}\frac{1}{(\frac{2z^2-2-3z}{iz})^{2}}\frac{dz}{iz}=$

$\displaystyle \displaystyle=\int_{0}^{2\pi}\frac{-z^{2}}{(2z^2-2-3z)^{2}}\frac{dz}{iz}=$

$\displaystyle \displaystyle=\int_{0}^{2\pi}\frac{-z}{(2z^2-2-3z)^{2}}\frac{dz}{i}$

How do I get rid off the minus in the numerator in the last integral? Is this correct so far? I need some help and suggestions to continue the solution.

When I solve for $\displaystyle z$ I get the following:

$\displaystyle \displaystyle{z_{1}=2$

$\displaystyle \displaystyle{z_{2}=-\frac{1}{2}$

I don't know if that is correct.

Thank you