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2.6 Q11


Yifei Wang:
Can someone help me with this question.
I am confused about this integral.

Victor Ivrii:
Start by writing the problem.

Amy Zhao:
Here is my solution´╝Ü
$I=\int_0^{2\pi} \frac{cos2\theta dz}{1-2acos\theta+a^2}~dx$
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, $\cos2\theta=\frac{1}{2} (z^2+z^{-2})$
I=&\int_{|z|=1}\frac{\frac{1}{2} (z^2+z^{-2})}{1-a(z+z^{-1})+a^2}\frac{dz}{iz}\\
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
 at $z=a$, $=-\frac{1+a^2}{a^2}$
By Residue Theorem, $I=2\pi i\frac{1}{2i}(\frac{a^4+1}{a^2(1-a^2)}-\frac{1+a^2}{a^2}=\frac{2\pi a^2}{1-a^2}$

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:
I think it is a typo.
The solution is as same as the derivative at z = 0.


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