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Homework Help: Trig Integral using Cauchy Formula

  1. Feb 27, 2009 #1
    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.
     
  2. jcsd
  3. Feb 28, 2009 #2

    CompuChip

    User Avatar
    Science Advisor
    Homework Helper

    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.
     
  4. Mar 1, 2009 #3
    Thank you. I totally missed the significance of [tex]0 < a < 1[/tex]. It makes a lot more sense now.
     
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