### Author Topic: 2.6 Q11  (Read 2395 times)

#### Yifei Wang

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##### 2.6 Q11
« on: November 19, 2018, 08:53:14 PM »
Can someone help me with this question.

#### Victor Ivrii

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##### Re: 2.6 Q11
« Reply #1 on: November 19, 2018, 09:22:00 PM »
Start by writing the problem.

#### Amy Zhao

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##### Re: 2.6 Q11
« Reply #2 on: November 19, 2018, 09:39:32 PM »
$\newcommand{\Res}{\operatorname{Res}}$
Here is my solution：
$I=\int_0^{2\pi} \frac{cos2\theta dz}{1-2acos\theta+a^2}~dx$
By substitution:   let $z=e^{i\theta}$, then $dz=ie^{i\theta}, d\theta= \frac{dz}{ie^{i\theta}}=\frac{dz}{iz}$
$cos\theta=\frac{e^{i\theta}+e^{-i\theta}}{2}=\frac{1}{2} (z+z^{-1})$
Similarly, $\cos2\theta=\frac{1}{2} (z^2+z^{-2})$
\begin{align*}
I=&\int_{|z|=1}\frac{\frac{1}{2} (z^2+z^{-2})}{1-a(z+z^{-1})+a^2}\frac{dz}{iz}\\
=&\frac{1}{2i}\int_{|z|=1}\frac{z^2+\frac{1}{z^2}}{z-az^2-a+a^2z}dz\\
=&\frac{1}{2i}\int_{|z|=1}\frac{z^4+1}{z^2(z-az-a+a^2z)}dz\\
=&\frac{1}{2i}\int_{|z|=1}\frac{z^4+1}{z^2(z-a)(1-az)}dz
\end{align*}
Let $f(z)=\frac{z^4+1}{z^2(z-a)(1-az)}$
$f(z)$ has simple pole at $z=a, z=\frac{1}{a}$, and pole of order 2 at $z=0$, but since $-1<a<1$, we only consider $z=a$ and $z=1$.z=0
\begin{align*}
&\Res(f(z);a)=\Res(\frac{\frac{z^4+1}{z^2(1-az)}}{z-a};a)=\frac{a^4+1}{a^2(1-a^2)},\\
&\Res(f(z);0)=\Res(\frac{\frac{z^4+1}{(z-a)(1-az)}}{(z-0)^2};0)=\frac{d}{dz}\frac{z^4+1}{(z-a)(1-az)}
\end{align*}
at $z=a$, $=-\frac{1+a^2}{a^2}$
By Residue Theorem, $I=2\pi i\frac{1}{2i}(\frac{a^4+1}{a^2(1-a^2)}-\frac{1+a^2}{a^2}=\frac{2\pi a^2}{1-a^2}$
« Last Edit: November 29, 2018, 05:19:52 AM by Victor Ivrii »

#### syee

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##### Re: 2.6 Q11
« Reply #3 on: November 22, 2018, 12:36:52 AM »
Sorry, I'm confused on what you did at the second last step. Why did you take the derivative of the residue at z=a?

#### Ruosi Ding

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##### Re: 2.6 Q11
« Reply #4 on: November 28, 2018, 10:26:53 PM »
I think it is a typo.
The solution is as same as the derivative at z = 0.