How would I compute:

$\displaystyle I=\int_{2\pi }^{0}\frac{4d\theta }{2 cos\theta + 3 sin\theta + 2i}$

I have tried finding the poles by changing the values into complex values

where $\displaystyle z=e^{i\theta }$

$\displaystyle 2cos\theta =z+\frac{1}{z}$

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

I have concluded

$\displaystyle I=\oint \frac{\frac{4dz}{iz}}{(z+\frac{1}{z})-\frac{3i}{2}(z-\frac{1}{z})+2i}$

and have found the poles:

$\displaystyle z_{1}=\frac{(12-8i)(2+\sqrt{7})}{52}$

and

$\displaystyle z_{2}=\frac{(12-8i)(2-\sqrt{7})}{52}$

Now how would I find out which one of the poles contributes and how do I continue to finish off the rest of the integral, thanks in advance..