1. ## residue problem

Hi there

I've been trying to revise and have come across a question in Schaum's Advanced Calculus that has done out a question but has skipped a lot of steps and so I can't completely follow what's going on. I'd be ever so grateful if anyone could fill in the gaps for me.

The question is:
Show that $\int_{0}^{2\pi}\frac{\cos3\theta \: d\theta}{5-4cos\theta}$= $\frac{pi}{12}$

I understand it all up to where he gets to

$\int_{0}^{2\pi}\frac{\cos3\theta \: d\theta}{5-4cos\theta}$= $\frac{-1}{2i}\int\frac{z^6+1}{z^3(2z-1)(z-2)}$
and I understand why he tries to find the residue at z=0 and $z=\frac{1}{2}$.

I'd really appreciate it if anyone could tell me step by step how when
z=0 the residue = $\frac{21}{8}$
and when $z=\frac{1}{2}$ the residue = $\frac{-65}{24}$

I'm new to these residue problems and am trying my best to work through them but skipping big steps in books doesn't help so if anyone could fill in the missing steps, I'd really appreciate it.

Regards
Michaela

$\int_0^{2\pi} \frac{e^{3i\theta}}{5-4\cos \theta} d\theta= \int_0^{2\pi}\frac{\cos 3\theta}{5-4\cos \theta} d\theta + i \int_0^{2\pi}\frac{\sin 3\theta }{5-4\cos \theta} d\theta$.
Note that,
$\int_0^{2\pi} \frac{e^{3i\theta}}{5-4\cos \theta} d\theta = \oint_{|z|=1} \frac{z^3}{5 - 2(z+z^{-1})}\cdot \frac{1}{iz} dz = \frac{1}{i} \oint_{|z|=1} \frac{z^3}{(1-2z)(z-2)} dz$.
By residue theorem,
$\oint_{|z|=1} \frac{z^3}{(1-2z)(z-2)} dz = \frac{2\pi i}{24} = \frac{\pi i}{12}$.
Thus,
$\int_0^{2\pi} \frac{e^{3 i \theta}}{5-4\cos \theta} d\theta = \frac{\pi i}{12} \implies \int_0^{2\pi} \frac{\cos 3\theta}{5-4\cos \theta} d\theta = \frac{\pi}{12}$.

3. Thank you so much for all the effort you went to to try and fill in the gaps but to be honest I still don't really know how to do it your way. I understood in the book where they got $\int_{0}^{2\pi}\frac{\cos3\theta \: d\theta}{5-4cos\theta}$= $\frac{-1}{2i}\int\frac{z^6+1}{z^3(2z-1)(z-2)}$ but it's the next couple of steps from this ie the parts when z=0 the residue = $\frac{21}{8}$ and when $z=\frac{1}{2}$ the residue = $\frac{-65}{24}$ that I'm stuck on.

I really do appreciate you trying to help me but I'm behind enough in this topic as it is without trying to deviate from the way the book does the questions.

I have no doubt that your method is probably more efficient ie when you got
$\oint_{|z|=1} \frac{z^3}{(1-2z)(z-2)} dz = \frac{2\pi i}{24} = \frac{\pi i}{12}$. Thus, $\int_0^{2\pi} \frac{e^{3 i \theta}}{5-4\cos \theta} d\theta = \frac{\pi i}{12} \implies \int_0^{2\pi} \frac{\cos 3\theta}{5-4\cos \theta} d\theta = \frac{\pi}{12}$.

I need to, however, keep to the way I've some idea of ie taking it from $\frac{-1}{2i}\int\frac{z^6+1}{z^3(2z-1)(z-2)}$ and then doing the residue part ie when z=0 the residue = $\frac{21}{8}$ and when $z=\frac{1}{2}$ the residue = $\frac{-65}{24}$

If you have the time and could show me how the book gets $\frac{21}{8}$ and $\frac{-65}{24}$, I'd be eternally grateful!

Thanks again for your help thus far, I really appreciate your time

Regards
Michaela

4. Originally Posted by michaela-donnelly
$\frac{-1}{2i}\int\frac{z^6+1}{z^3(2z-1)(z-2)}$ and then doing the residue part ie when z=0 the residue = $\frac{21}{8}$ and when $z=\frac{1}{2}$ the residue = $\frac{-65}{24}$
I assume you are integrating over the unit circle. You forgot to state what you are integrating.
Thus, I will assume it is,
$\oint \limits_{|z|=1}\frac{z^6+1}{z^3(2z-1)(z-2)} dz$.

The function,
$f(z) = \frac{z^6+1}{z^3(2z-1)(z-2)}$.
Has poles at $z=0,z=1/2,z=2$, but the only ones inside the circle $|z|=1$ are $z=0,1/2$.

We note that $z=1/2$ is a simple pole thus,
$\mbox{res}(f,1/2) = \lim_{z\to 1/2} (z-1/2) \cdot \frac{z^6+1}{(2z-1)(z-2)} = \lim_{z\to 1/2} (2z-1)\frac{\frac{1}{2}z^6+\frac{1}{2}}{(2z-1)(z-2)} = ?$

The real bad point is at $z=0$ it is a triple pole thus,
$\mbox{res}(f,0) = \lim_{z\to 0} \frac{1}{(3-1)!}\left[ z^3\cdot \frac{z^6+1}{z^3(2z-1)(z-2)} \right]'' = ?$

You do those calculations and you get your answer after you add up those residues. Since you are doing complex integration maybe this will help? It is not complete yet, I do not have time to work on it now.

5. ## residue problem

Thank you SO much for your help. I'll work through that and see how I get on. I really appreciate you taking the time to help me out.

thanks again

6. ## residue problem

Hi all

I've tried to work through the previous solution but still can't fill in all the gaps! Can anyone explain it to me using my method???

ANY suggestions would be a great help

Michaela