Trig Integral using Cauchy Formula

[Bacat]

Homework Statement

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)

Homework 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)

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[/color] \frac{1}{a} is outside the circle[/color] |z| = 1, and the point a is inside the circle[/color] |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,[/color]

\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.

 

[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&#039;(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.

 

[Bacat]

Thank you. I totally missed the significance of 0 &lt; a &lt; 1. It makes a lot more sense now.