# Find the value of $~\int_0^{2\pi}\frac{d\theta}{1-2a\cos\theta + a^2}~~~$ for $~|a|<1~.$ [duplicate]

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.

• 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. – nmasanta Aug 24 at 15:18
• The only singularity in the unit disk is in $a$. @nmasanta – Surb Aug 24 at 15:56
• 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 ? – nmasanta Aug 24 at 16:24
• 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... – Surb Aug 24 at 16:38
• Yes, that's what I mean. Thanks – nmasanta Aug 24 at 16:51