# Homework Help: Trig Integral using Cauchy Formula

1. Feb 27, 2009

### Bacat

1. The problem statement, all variables and given/known data

Evaluate the integral:

$$I=\frac{1}{2\pi} \int^{2\pi}_0 \frac{d\theta}{1-2aCos\theta + a^2}, 0 < a < 1.$$

This integral is worked out in the book as an example, but I don't understand all the steps. My confusion is highlighted in red below.

(From Complex Variables, Stephen Fisher (2nd Edition), Chapter 2.3, Exercise 7)

2. Relevant equations

$$\frac{1}{2\pi i}\oint_{\gamma} \frac{f(s)}{s-z} ds= \left\{ \begin{array}{lr}f(z) & : z \in \gamma\\0 & : z \notin \gamma \end{array}$$ where $$\gamma$$ is defined as the interior of the simple closed curve described by the line integral. (Eq. 1)

$$Cos\theta=\frac{1}{2}\left(z+\frac{1}{z}\right).\;$$ (Eq. 2)

$$d\theta = \frac{dz}{iz}. \;$$ (Eq. 3)

3. The attempt at a solution

Substitute using (2)...

$$1-2aCos\theta + a^2=1+a^2-a\left(z + \frac{1}{z}\right).$$

Now using (3)...

$$\frac{d\theta}{1-2aCos\theta + a^2} = \frac{dz}{i(-az^2+(1+a^2)z-a)} \!$$ (Eq. 4)

Now:

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

The point $$\frac{1}{a}$$ is outside the circle $$|z| = 1$$, and the point a is inside the circle $$|z|=1$$.

How do we know that 1/a is outside and a is inside?

Hence,

$$\frac{1}{2\pi i} \oint_{|z|=1} \frac{dz}{-az^2+(1+a^2)z-a}=\frac{1}{2\pi i}\oint_{|z|=1}\frac{1}{-a(z-\frac{1}{a})}\frac{1}{z-a} dz = \frac{1}{-a\left(a-\frac{1}{a}\right)} = \frac{1}{1-a^2}$$

But,

$$\frac{1}{2\pi i} \oint_{|z|=1} \frac{dz}{-az^2+(1+a^2)z-a}=\frac{1}{2\pi i} \int_0^{2\pi} \frac{i e^{i\theta}d\theta}{e^{i\theta}(1-2aCos\theta + a^2)}$$

Why do we need this equality? I thought we already substituted Eqs. (2) and (3) to get expression (4). In other words, why are we bothering to put exponentials in here when we already have a simple form for the integral?

$$=\frac{1}{2\pi} \int_0^{2\pi} \frac{d\theta}{1-2aCos\theta +a^2}.$$

Thus, the integral has the value

$$\frac{1}{2\pi} \int_0^{2\pi} \frac{d\theta}{1-2aCos\theta +a^2} = \frac{1}{1-a^2}, \; 0 < a < 1.$$

Thank you for your time. I know this is a lot of equations to look through, but it would really help this confused student who wants to understand the example.

2. Feb 28, 2009

### CompuChip

1)
Well, it is given that 0 < a < 1, you wrote that on the very first line. So |a| < 1. Then |1/a| = 1/|a| > 1. Geometrically, the reciprocal of any complex number is obtained by reflecting the number in the unit circle. If you want to show it explicitly, do something like |1/a| * |a| = |1/a * a| = |1| = 1.

2)
You have something like
$$\oint_{|z| = 1} f(z) \, dz$$
which is a meaningful expression in terms of concepts, but it's useless in terms of calculation. The contour integral of f(z) over a closed curve $\gamma: t \mapsto \gamma(t) \in \mathbb{C}$ defined on [a, b] is defined (?) as
$$\oint_\gamma f(z) \, dz = \int_a^b f(\gamma(t)) \gamma'(t) \, dt$$.
So if you want to evaluate your contour integral over the circle |z| = 1 you will need some curve parametrizing it, and
$$\gamma: [0, 2\pi] \to \{ z \in \mathrm{C} : |z| = 1 \}, t \mapsto e^{\imi t}$$
is the most obvious one.

However, in this case you can also use calculus of residues, to immediately evaluate
$$\frac{1}{2\pi i}\oint_{|z|=1}\frac{1}{-a(z-\frac{1}{a})}\frac{1}{z-a} dz = \frac{1}{-a\left(a-\frac{1}{a}\right)} = \frac{1}{1-a^2}$$
without using an explicit parametrization.

Hope that helps.

3. Mar 1, 2009

### Bacat

Thank you. I totally missed the significance of $$0 < a < 1$$. It makes a lot more sense now.