[Complex Analysis] Help with Cauchy Integral Problem

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Homework Statement



Evaluate the following integral,

I = \int_{0}^{2\pi} \frac{d \theta}{(1-2acos \theta + a^2)^2}, \ 0 < a < 1

For such, transform the integral above into a complex integral of the form ∫Rₐ(z)dz, where Rₐ(z) is a rational function of z. This will be obtained through the substitution z = e^iθ. Therefore, solve the complex integral through Cauchy's Integral Formula for the first derivative of an analytic function.

Homework Equations



Cauchy's Integral Formula

f^{(n)} (z_0) = \frac{n!}{2 \pi i} \oint_{\gamma} \frac{f(z)}{(z - z_0)^{(n+1)}}dz

The Attempt at a Solution



Writting dθ in terms of dz

z = e^{i \theta}

dz = ie^{i \theta} d\theta

\frac{1}{iz}dz = d\theta

Writting

\frac{1}{(1-2acos \theta + a^2)^2}

in terms of z, known that

\frac{1}{2}\left( z + \frac{1}{z}\right) = cos \theta

Therefore

\frac{1}{(1-a\left( z + \frac{1}{z}\right) + a^2)^2}

Thus

I = \oint_{\gamma} \frac{dz}{iz(1-a\left( z + \frac{1}{z}\right) + a^2)^2}

Where do I go from here? Any clues on how to express the denominator in terms of (z-z0)² ?
 
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Try multiplying by z/z. You will have a factor of ##z^2## in the denominator. Thus,

##z^2 \cdot (1-a(z+z^{-1})+a^2)^2 = [z \cdot (1-a(z+z^{-1})+a^2)]^2##

It should be apparent from there!

Edit: When you finally do this, it should become apparent why ##0 < a < 1## is required.
 
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