Using residue theorem, find the value of $$~\int_0^{2\pi}\dfrac{d\theta}{1-2a\cos\theta + a^2}$$ for $~|a|<1~.$

I know that the value of the integral is $~\frac{2\pi}{1-a^2}~$(I found it by using the rule of normal definite integral), but don't know how to find the same by using residue theorem.

Any one please provide me a complete solution by using residue theorem.


Hint $$\cos\theta =\frac{e^{i\theta }+e^{-i\theta }}{2}.$$ Thus, making the substitution $z=e^{i\theta }$ yields $$\int_0^{2\pi}\frac{\,\mathrm d \theta }{1-2a\cos(\theta )+a^2}=\int_{\gamma }\frac{z}{z-a(z^2+1)+za^2}\cdot \frac{1}{iz}\,\mathrm d z,$$ where $\gamma $ is the unit circle.

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  • $\begingroup$ Thanks for your response. I try this and got two point $~z=a,~~1/a~$, which are both simple pole. I found residue at these simple poles. Now when using residue theorem I got the value is $~0~$. I can't understand where I am wrong. Please help. $\endgroup$ – nmasanta Aug 24 at 15:18
  • $\begingroup$ The only singularity in the unit disk is in $a$. @nmasanta $\endgroup$ – Surb Aug 24 at 15:56
  • $\begingroup$ How to show that $~z=1/a~$ is in the outside of the unit disk ?$ ~|z|=|1/a|=1\implies |a|=1~$. But given that $|a|<1$. Is it okay ? $\endgroup$ – nmasanta Aug 24 at 16:24
  • $\begingroup$ Not clear what you are doing. The unit disk is $\{z\in\mathbb C\mid |z|<1\}$. So, if $|a|<1$, obviously $\frac{1}{|a|}>1$ and thus $\frac{1}{a}$ is clearly not in the unit disk... $\endgroup$ – Surb Aug 24 at 16:38
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    $\begingroup$ Yes, that's what I mean. Thanks $\endgroup$ – nmasanta Aug 24 at 16:51

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