Sorry, I'm new at posting here, so forgive me for any mistakes I make.
I'm trying to evaluate the following using a contour integral. I don't know how to use the Residue stuff yet, so I basically just have to change the integral to be in terms of z, break the integrand up into partials fractions, and solve by applying Cauchy's Theorem (i.e., the whole if the root that causes discontinuity exists outside of the closed curve, then that integral evaluates to zero thing, roughly). Here's the problem in question:
$$ \int_0^{2\pi}\frac{\mathrm{d}\theta}{3-2\cos{\theta}+\sin{\theta}} $$
I get the part where I have to substitute in z (i.e., $z=e^{i\theta}=\cos{\theta}+i\sin{\theta}$ and what not), and solving for the roots. I got to here:
$$ \int_\gamma\frac{2\mathrm{d}z}{(1 - 2i)z^2 + 6iz -1-2i} $$
And using the quadratic formula, I found the roots to be $2 - i$ and $\frac{2-i}{5}$. So then the above integral should equal the following, right?
$$ \int_\gamma\frac{2\mathrm{d}z}{(z-2+i)(z-\frac{2-i}{5})} $$
I know something's wrong here because those two integrals don't equal each other. It's probably due to the lack of coffee, but whatever. If I try partial fractions, I get $A = \frac{2+i}{2}$ and $B = \frac{-4-3i}{5}$, so my final integral that I'm supposed to evaluate is something like:
$$ \frac{2+i}{2}\int_\gamma\frac{\mathrm{d}z}{z-2+i} + \frac{-4-3i}{5}\int_\gamma\frac{\mathrm{d}z}{z-\frac{2-i}{5}} $$
...Except not wrong, obviously, because this is supposed to evaluate to $\pi$. Major questions:
- $\gamma$ is supposed to be the closed unit circle, so would I just evaluate this integral from 0 to 1?
- And what would I do if I wanted to evaluate it on say, a rectangle instead?
You forgot to multiply your factored polynomial by the leading coefficient $1-2i$ which of course doesn't happen automatically: $$(1−2i)z^2+6iz−1−2i=(1-2i)(z−2+i)(z−\frac{2−i}{5})$$
Contour integration in this case simplifies the solution because of the Residue theorem: integrating the function $\oint \frac{dz}{z-a}$ gives $2\pi i$ if the pole $a$ is within the contour, and $0$ otherwise. So there isn't really anything to integrate, just check the position of the poles and you are done. If your contour were not a circle, just modify the test which poles are inside the contour.