
Complex Integration II
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:
$\displaystyle z_{1}=\frac{(128i)(2+\sqrt{7})}{52}$
and
$\displaystyle z_{2}=\frac{(128i)(2\sqrt{7})}{52}$
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..

When I just integrate it from 0 to 2pi and make all the subs I get:
$\displaystyle 4i\oint \frac{2i}{2i(z^2+1)+3(z^21)4z}dz$
lil' of this, lil' of that and I get:
$\displaystyle \frac{8}{2i+3}\oint \frac{1}{z^2\frac{4}{2i+3}z+\frac{2i3}{2i+3}}dz$
$\displaystyle \frac{8}{2i+3}\oint \frac{1}{(z\alpha)(z\beta)}dz$
and only one is in the contour (via Mathematica Abs command) giving a residue of $\displaystyle (3/2+i)\sqrt{17}$
then I get:
$\displaystyle 2\pi i\frac{8}{2i+3}(3/2+i)/\sqrt{17}\approx 6.1 i$ which agrees with the numerical result. Yours is the negative of that.