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

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..