[Complex Analysis] Help with Cauchy Integral Problem

[Je m'appelle]

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 ?

 

[scurty]

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.