Trig Integral using Cauchy Formula

[Bacat]

Homework Statement

Evaluate the integral:

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

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

[tex]\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} [/tex] where [tex]\gamma[/tex] is defined as the interior of the simple closed curve described by the line integral. (Eq. 1)

[tex]Cos\theta=\frac{1}{2}\left(z+\frac{1}{z}\right).\;[/tex] (Eq. 2)

[tex]d\theta = \frac{dz}{iz}. \;[/tex] (Eq. 3)

The Attempt at a Solution

Substitute using (2)...

[tex]1-2aCos\theta + a^2=1+a^2-a\left(z + \frac{1}{z}\right).[/tex]

Now using (3)...

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

Now:

[tex]-az^2+(1+a^2)z-a=-a\left(z-\frac{1}{a}\right)(z-a)[/tex]

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

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

Hence,

[tex]\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}[/tex]

But,

[tex]\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)}[/tex]

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?

[tex]=\frac{1}{2\pi} \int_0^{2\pi} \frac{d\theta}{1-2aCos\theta +a^2}.[/tex]

Thus, the integral has the value

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

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
[tex]\oint_{|z| = 1} f(z) \, dz[/tex]
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 [itex]\gamma: t \mapsto \gamma(t) \in \mathbb{C}[/itex] defined on [a, b] is defined (?) as
[tex]\oint_\gamma f(z) \, dz = \int_a^b f(\gamma(t)) \gamma'(t) \, dt[/tex].
So if you want to evaluate your contour integral over the circle |z| = 1 you will need some curve parametrizing it, and
[tex]\gamma: [0, 2\pi] \to \{ z \in \mathrm{C} : |z| = 1 \}, t \mapsto e^{\imi t}[/tex]
is the most obvious one.

However, in this case you can also use calculus of residues, to immediately evaluate
[tex]\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}[/tex]
without using an explicit parametrization.

Hope that helps.

 

[Bacat]

Thank you. I totally missed the significance of [tex]0 < a < 1[/tex]. It makes a lot more sense now.