(adsbygoogle = window.adsbygoogle || []).push({}); 1. The problem statement, all variables and given/known data

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)

2. Relevant 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)

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.

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# Homework Help: Trig Integral using Cauchy Formula

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