MHB Integration using residue theorem

[aruwin]

Hi. I have to use the residue theorem to integrate f(z).
Can someone help me out? I am stuck on the factorization part.

Find the integral
$$\int_{0}^{2\pi} \,\frac{d\theta}{25-24\cos\left({\theta}\right)}$$

My answer:
$$\int_{0}^{2\pi} \,\frac{d\theta}{25-24\cos\left({\theta}\right)}=\oint_{c}^{} \,\frac{dz/iz}{25-24(\frac{1}{2}(z+\frac{1}{z}))}$$

$$=\frac{1}{i}\oint_{c}^{} \,\frac{dz}{-12z^2+25z-12}$$

 

[chisigma]

Hi. I have to use the residue theorem to integrate f(z).
Can someone help me out? I am stuck on the factorization part.

Find the integral
$$\int_{0}^{2\pi} \,\frac{d\theta}{25-24\cos\left({\theta}\right)}$$

My answer:
$$\int_{0}^{2\pi} \,\frac{d\theta}{25-24\cos\left({\theta}\right)}=\oint_{c}^{} \,\frac{dz/iz}{25-24(\frac{1}{2}(z+\frac{1}{z}))}$$

$$=\frac{1}{i}\oint_{c}^{} \,\frac{dz}{-12z^2+25z-12}$$

Excellent!... now find the poles of $\displaystyle f(z) = - \frac{1}{i\ (12\ z^{2} - 25\ z + 12)}$ and then compute the residues of the poles inside the unit circle...

Kind regards

$\chi$ $\sigma$

 

[aruwin]

Here's my continuation:

$$\frac{1}{i}\oint_{c}^{}\frac{dz}{-(4z-3)(3z-4)}=-\frac{1}{i}\oint_{c}^{}\frac{dz}{(4z-3)(3z-4)}$$

It has a singularity at z=3/4

$$Resf(z)_{|z=\frac{3}{4}|}=\lim_{{z}\to{\frac{3}{4}}}(z-\frac{3}{4})(\frac{1}{(4z-3)(3z-4)})=\frac{-7}{5}$$

By residue theorem, the integral becomes
$$2\pi{i}\frac{-1}{i}\frac{-7}{5}=\frac{14\pi}{5} $$

Is this correct? Check please.

 

[chisigma]

Here's my continuation:

$$\frac{1}{i}\oint_{c}^{}\frac{dz}{-(4z-3)(3z-4)}=-\frac{1}{i}\oint_{c}^{}\frac{dz}{(4z-3)(3z-4)}$$

It has a singularity at z=3/4

$$Resf(z)_{|z=\frac{3}{4}|}=\lim_{{z}\to{\frac{3}{4}}}(z-\frac{3}{4})(\frac{1}{(4z-3)(3z-4)})=\frac{-7}{5}$$

By residue theorem, the integral becomes
$$2\pi{i}\frac{-1}{i}\frac{-7}{5}=\frac{14\pi}{5} $$

Is this correct? Check please.

Excellent!...

Kind regards

$\chi$ $\sigma$

 

[aruwin]

Excellent!...

Kind regards

$\chi$ $\sigma$

I had a miscalculation. The final answer should be $\frac{2\pi}{7}$.

 

[chisigma]

I had a miscalculation. The final answer should be $\frac{2\pi}{7}$.

Effectively there was a mistake in the computation of the residue...

$\displaystyle r = - \frac{1}{i}\ \lim_{z \rightarrow \frac{3}{4}} \frac{z - \frac{3}{4}}{(3 z - 4)\ (4 z - 3)} = \frac{1}{7\ i}$

Kind regards

$\chi$ $\sigma$