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Purely imaginary complex contour

  1. Nov 8, 2005 #1

    saltydog

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    I've been working with Complex Analysis and have noticed an interesting result.

    Under what conditions will the following integral be purely imaginary:

    [tex]\int_{a-bi}^{a+bi} f(z)dz[/tex]

    It seems to me some type of symmetry is required. Take for example:

    [tex]\int_{1-8i}^{1+8i} f(z)dz[/tex]

    where:

    [tex]f(z)= \frac{e^{3z}}{z}[/tex]

    Now:

    [tex]\int_{1-8i}^{1+0i} \frac{e^{3z}}{z}dz\approx 10.6559+2.7271i[/tex]

    and:

    [tex]\int_{1+0i}^{1+8i} \frac{e^{3z}}{z}dz\approx -10.6559 + 2.7271i[/tex]

    Thus:

    [tex]\int_{1-8i}^{1+8i} \frac{e^{3z}}{z}dz\approx 5.4543i[/tex]

    Note the plot below which is the image of f(z) in the u-v plane for f(z) along the path indicated above. Red is the image for negative values of b; blue is for positive values. The symmetry is obvious.

    I've noticed this with other similar integrals and suspect symmetry causes the integral to be purely imaginary but haven't yet figured out how to prove it. Can anyone offer ideas/hints/suggestions?
     

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    Last edited: Nov 9, 2005
  2. jcsd
  3. Nov 9, 2005 #2

    saltydog

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    Alright, I'm making some progress in this matter:

    Consider the problem:

    [tex]\int_{1-8i}^{1+8i} \frac{e^z}{z}dz[/tex]

    Expanding the integrand into a Real part and Imaginary part:

    [tex]
    \begin{align*}
    \frac{e^{x+yi}}{(x+yi)}&=\frac{e^x}{x^2+y^2}\left(xCos(y)+ ySin(y)\right) \\
    &+ i\left[\frac{e^x}{x^2+y^2}\left(yCos(y)+xSin(y)\right)\right] \\
    &=u+vi
    \end{align}
    [/tex]

    and using the relation:

    [tex]\int_c f(z)dz=\int_c udx-vdy+i\int_c udy+vdx[/tex]

    I let:

    [tex]x=1,\quad dx=0[/tex]

    [tex]y=t,\quad dy=dt[/tex]

    The integral then becomes:

    [tex]
    \begin{align*}
    \int_{1-8i}^{1+8i} \frac{e^z}{z}dz&=-k\int_{-8}^8 \left(\frac{tCos(t)}{1+t^2}+\frac{Sin(t)}
    {1+t^2}\right)dt \\
    &+i\left(\int_{-8}^{8} \left(\frac{Cos(t)}{1+t^2}+\frac{tSin(t)}{1+t^2}\right)dt
    \end{align}
    [/tex]

    Note that the real part is an odd function of t and thus the integral over a symmetric region about the origin is zero. Thus the real part of the integral is zero. The imaginary part however is NOT and odd function.

    Thus my first conclusion is:

    If the Real part of the Euler expansion of the integrand is an odd function of the parameter t, then the results will be either zero or pure imaginary.
     
    Last edited: Nov 9, 2005
  4. Nov 9, 2005 #3

    Galileo

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    Interesting. Going from 1-8i to 1 gives you the same answer as from 1 to 1+8i, but with a flip in the real part.
    Actually, it just occured that it is a trivial consequence of the reflection principle. If an analytic function f(z) is defined on a region that is symmetric wrt the x-axis, contains the x-axis and takes on real values on that axis. Then
    [tex]\bar f(z)=f(\bar z)[/tex].
     
  5. Nov 9, 2005 #4

    saltydog

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    Thanks Galileo. I'll look that up. Might you comment why inverse Laplace Transforms have this property? That is the reason of course that when the integral expression for the inverse transform is multiplied by the coefficient:

    [tex]\frac{1}{2\pi i}[/tex]

    a function of a real variable is obtained.:smile:
     
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