1. ## complex analysis

$I=\int^{2\pi}_{0}\frac{cos(3\theta)}{5-4\cos(\theta)}d\theta$

$z=e^{i\theta},\frac{dz}{d\theta}=i\\e^{i\theta}=i\ \z\Rightarrow\\d\theta=\frac{dz}{i\\z}$

$I=\oint_{C_{1}(0)}\frac{z^{3}+z^{-3}}{2\\i(-2z^{2}+5z-2)}dz$

factorizing gives

$I=\oint_{C_{1}(0)}\frac{z^{3}+z^{-3}}{2\\i(z-2)(z-\frac{1}{2})}dz$

now from here.. i need to do $2\pi\\i*\sum\\Res\\|_{z}$ but everytime i do it, i get dumb answers, im pretty sure the answer is $\frac{\pi}{12}$

please can anyone help me.. i may have done something stupid

2. I think the residues are at z=0,1/2 and 2... but not sure... and the contour only encloses z=0 and 1/2

3. Originally Posted by fredrick08
$I=\int^{2\pi}_{0}\frac{cos(3\theta)}{5-4\cos(\theta)}d\theta$
Rewrite it as $iI=\int_0^{2\pi}\frac{\left(e^{i\theta}\right)^2}{ 5-4\frac{e^{i\theta}+\frac{1}{e^{i\theta}}}{2}}e^{i\ theta}d\theta$ but this is clearly $\underset{|z|=1}{\oint}\frac{z^2}{5-2(z+\frac{1}{z})}dz=\underset{|z|=1}{\oint}\frac{-z^3}{2(z-2)(z-\tfrac{1}{2})}$. But, clearly $f(z)=\frac{-z^3}{2(z-2)}$ is analytic on $\mathbb{D}$ and so Cauchy's integral formula tells us that $iI=2\pi i\frac{-(\tfrac{1}{2})^3}{2(\tfrac{1}{2}-2)}=\frac{\pi i}{12}\implies I=\frac{\pi}{12}$

.

4. ok yes, that makes perfect sense... but could please do it by residue theorem, coz thats how we are meant to atm.. srry but also.... how did u get the -z^3 on top.... where does the z^-3 go?

because i know that $z^{3}+z^{-3}=\frac{z^{6}+1}{z^{3}}$ can that work at all?

5. Originally Posted by fredrick08
ok yes, that makes perfect sense... but could please do it by residue theorem, coz thats how we are meant to atm
.........

Ok, well let $f(z)$ be as above and note that by it's analyticity that $f(z)=\sum_{n=0}^{\infty}\frac{f^{(n)}(\tfrac{1}{2} )(z-\tfrac{1}{2})^n}{n!}$ and so $\frac{f(z)}{z-\tfrac{1}{2}}=\frac{f(\tfrac{1}{2})}{z-\tfrac{1}{2}}+\frac{1}{2}f'(\tfrac{1}{2})(z-\tfrac{1}{2})+\cdots$ and so $\underset{z=\tfrac{1}{2}}{\text{Res}}\text{ }\frac{f(z)}{z-\tfrac{1}{2}}=f(\tfrac{1}{2})$..........

6. Originally Posted by fredrick08
-z^3 on top.... where does the z^-3 go?
$\frac{z^2}{5-2(z+\tfrac{1}{z})}=\frac{z^3}{5z-2(z^2+1)}=\frac{z^3}{-2z^2+5z-2}=\frac{-z^3}{2z^2-5z+2}$

7. Originally Posted by Drexel28
$\frac{z^2}{5-2(z+\tfrac{1}{z})}=\frac{z^3}{5z-2(z^2+1)}=\frac{z^3}{-2z^2+5z-2}=\frac{-z^3}{2z^2-5z+2}$
srry i dont understand how u went from

$\int^{2\pi}_{0}\frac{cos(3\theta)}{5-4\cos(\theta)}d\theta$

$iI=\int_0^{2\pi}\frac{\left(e^{i\theta}\right)^2}{ 5-4\frac{e^{i\theta}+\frac{1}{e^{i\theta}}}{2}}e^{i\ theta}d\theta$

how can u change cos(3theta) which is 1/2(z^3+z^-3) to z^2???

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8. Originally Posted by fredrick08
srry i dont understand how u went from

$\int^{2\pi}_{0}\frac{cos(3\theta)}{5-4\cos(\theta)}d\theta$

$iI=\int_0^{2\pi}\frac{\left(e^{i\theta}\right)^2}{ 5-4\frac{e^{i\theta}+\frac{1}{e^{i\theta}}}{2}}e^{i\ theta}d\theta$

how can u change cos(3theta) which is 1/2(z^3+z^-3) to z^2???

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Oops, forgot to explain that. I'm kinda used to just writing stuff. I'll let you explain, think about $\int_0^{2\pi}\frac{e^{3i\theta}}{5-4\cos(\theta)}d\theta$ and how that relates to the real part of the residue? It just happens that the real part of the residue is the entire residue and so I just omitted it. My sloppy mistake

9. Originally Posted by Drexel28
Oops, forgot to explain that. I'm kinda used to just writing stuff. I'll let you explain, think about

$\int_0^{2\pi}\frac{e^{3i\theta}}{5-4\cos(\theta)}d\theta$

and how that relates to the real part of the residue? It just happens that the real part of the residue is the entire residue and so I just omitted it. My sloppy mistake
srry i dont know what you mean by real part of residue??

are u saying that $\cos(3\theta)=\frac{1}{2}(z^{3}+z^{-3})=z^{3}$????

na i definately dont understand that... and is that how it became z^2.. because u brought i out the front then cancelled the z form i*z?

10. Originally Posted by fredrick08
srry i dont know what you mean by real part of residue??

are u saying that $\cos(3\theta)=\frac{1}{2}(z^{3}+z^{-3})=z^{3}$????

na i definately dont understand that... and is that how it became z^2.. because u brought i out the front then cancelled the z form i*z?
Ok, well I'll just say it then.

$\cos(3\theta)=\text{Re }e^{3i\theta}$...so?

11. Originally Posted by Drexel28
Ok, well I'll just say it then.

$\cos(3\theta)=\text{Re }e^{3i\theta}$...so?
hmm ya i see how the real part of e^i3(theta)=cos(3(theta) but how do u get z^2 form there? like

$e^{3\\i\theta}=\cos(3\theta)+i\sin(3\theta)=z^{2}$???? srry... i dont understand, my maths knowledge is not great..

12. Originally Posted by fredrick08
hmm ya i see how the real part of e^i3(theta)=cos(3(theta) but how do u get z^2 form there? like

$e^{3\\i\theta}=\cos(3\theta)+i\sin(3\theta)=z^{2}$???? srry... i dont understand, my maths knowledge is not great..
In some sense what we're saying is that $\int_0^{2\pi}\frac{\cos(3\theta)}{5-4\cos(\theta)}d\theta=\int_0^{2\pi}\text{Re }\frac{e^{3i\theta}}{5-4\cos(\theta)}d\theta=\text{Re }\int_0^{2\pi}\frac{e^{3i\theta}}{5-4\cos(\theta)}d\theta$ and the calculation I did shows that $\int_0^{2\pi}\frac{e^{3i\theta}}{5-4\cos(\theta)}d\theta=\frac{\pi}{12}$ and so our integral is equal to $\text{Re }\frac{\pi}{12}=\frac{\pi}{12}$