# complex analysis

#### fredrick08

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

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

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

factorizing gives

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

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

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

#### fredrick08

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

#### Drexel28

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

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#### fredrick08

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

#### Drexel28

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

#### Drexel28

MHF Hall of Honor
-z^3 on top.... where does the z^-3 go?
$$\displaystyle \frac{z^2}{5-2(z+\tfrac{1}{z})}=\frac{z^3}{5z-2(z^2+1)}=\frac{z^3}{-2z^2+5z-2}=\frac{-z^3}{2z^2-5z+2}$$

#### fredrick08

$$\displaystyle \frac{z^2}{5-2(z+\tfrac{1}{z})}=\frac{z^3}{5z-2(z^2+1)}=\frac{z^3}{-2z^2+5z-2}=\frac{-z^3}{2z^2-5z+2}$$
srry i dont understand how u went from

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

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

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

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#### Drexel28

MHF Hall of Honor
srry i dont understand how u went from

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

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

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

fredrick08

#### fredrick08

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

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

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 $$\displaystyle \cos(3\theta)=\frac{1}{2}(z^{3}+z^{-3})=z^{3}$$????

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?

#### Drexel28

MHF Hall of Honor
srry i dont know what you mean by real part of residue??

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

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.

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