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Complex Analysis: Harmonic Conjugates

  1. Aug 22, 2010 #1
    1. The problem statement, all variables and given/known data
    For [tex]u(x,y)=e^{-y}(x\sin(x)+y\cos(x))[/tex] find a harmonic conjugate [tex]v(x,y)[/tex] and express the analytic function [tex]f=u +iv[/tex] as a function of z alone (where [tex]z=x+iy[/tex]0

    2. Relevant equations
    The Cauchy Riemann equations [tex]u_x=v_y[/tex] and [tex]u_y=-v_x[/tex]
    and possibly:
    [tex]sin(x) = \frac{e^{ix}-e^{-ix}}{2i}[/tex]
    [tex]cos(x) = \frac{e^{ix}+e^{-ix}}{2}[/tex]

    3. The attempt at a solution
    I've been toying around with algebra and I've been able to reduce up to [tex]u=\frac {e^{-y+ix}(y-ix)+e^{-y-ix}(y+ix)}{2}[/tex] but I can't get past here. I have also noted that if I divide by i, u(x,y) becomes u(y,-x) so I can replace these variables and I have [tex] u/i = e^{x}(y\sin(y)-x\cos(y)[/tex] which is, as someone on another forum pointed out, the real part of [tex]-ze^{z}[/tex]. So I suppose I could write [tex]f=Re(-ze^{iz})i + iv[/tex] no?
    But how should I go about finding v? I am familiar with the straightforward procedure of finding u_x, integrating w.r.t v, then finding v_x and setting it equal to u_y... but I can't seem to simplify u to the point of easy differentiation. I don't know if I should just try to brute force it, or if that's even possible. It seems like u should simplify more.
  2. jcsd
  3. Aug 22, 2010 #2


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    I would not change the sine and cosine to exponentials, although you certainly can do so.

    With [itex]v(x,y)= e^{-y}(x sin(x)- y cos(x))[/itex], [itex]v_x= e^{-y}(sin(x)+ y cos(x))+ y sin(x))[/itex] and [itex]v_y= -e^{-y}(x sin(x)- y cos(x))+ e^{-y}(-cos(x))[/itex]

    Now you must find u(x,y) such that
    [tex]u_x= v_y= -e^{-y}(x sin(x)- y cos(x))+ e^{-y}(-cos(x))[/tex]
    [tex]u_y= -v_x= -e^{-y}(sin(x)+ y cos(x))- y sin(x))[/itex]

    Integrate either of those equations with respect to the appropriate variable, remembering that the "constant" of integration may be a function of the other variable, then put that into the other equation to find that function.
  4. Aug 22, 2010 #3
    Great. I guess the integration wasn't as bad as I thought. Your algebra is a bit off because you wrote down a different u(x,y) than I had.

    Using [tex]u(x,y)=e^{-y}(x\sin(x)+y\cos(x)[/tex] I came up with
    v(x,y) = e^{-y}(ysin(x)-xcos(x)).

    Now my question is how to put this in terms of z only. I think I have a THM I can use: http://planetmath.org/encyclopedia/IdentityTheoremOfHolomorphicFunctions.html" [Broken]. So if I have f(u,v) = e^{-y}(xsin(x)+ycos(x) +i(e^{-y}(ysin(x)-xcos(x)) can I say that some function g(z) is equal to f(z) along the iy axis of the complex plane (i.e. where x=o). Then f(z)=g(z) for all complex numbers. So f(z)=g(0+iy) ---> f(z)=ye^{-y}?

    Sorry about the formatting, Latex is giving me trouble.
    Last edited by a moderator: May 4, 2017
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