Complex Integration II

Apr 2010
26
1
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..
 

shawsend

MHF Hall of Honor
Aug 2008
903
379
When I just integrate it from 0 to 2pi and make all the subs I get:

\(\displaystyle -4i\oint \frac{2i}{2i(z^2+1)+3(z^2-1)-4z}dz\)

lil' of this, lil' of that and I get:

\(\displaystyle \frac{8}{2i+3}\oint \frac{1}{z^2-\frac{4}{2i+3}z+\frac{2i-3}{2i+3}}dz\)

\(\displaystyle \frac{8}{2i+3}\oint \frac{1}{(z-\alpha)(z-\beta)}dz\)

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.