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Homework Help: Complex Analysis - Harmonic Functions

  1. Sep 18, 2012 #1
    1. The problem statement, all variables and given/known data

    Show ln(az) where a is a real number and z = x + iy is harmonic everywhere except z = 0.

    2. Relevant equations

    z = x + iy = rcos(θ) + irsin(θ) = re^iθ
    z = u(x,y) + iv(x,y)

    Cauchy Riemann test for analyticity:
    ∂u/∂x = ∂v/∂y
    ∂u/∂y = -∂v/∂x

    3. The attempt at a solution

    ln(az) = ln (rcos(θ) + irsin(θ)) = ln(rcos(θ+2nπ) + irsin(θ+2nπ))

    = ln(a*re^iθ) = ln(ar) + i(θ+2nπ) <- this is multivalued, not harmonic.

    How do I show it is harmonic?
  2. jcsd
  3. Sep 18, 2012 #2
    Either rewrite ln(az) using x and y or the C-R relations using r and θ (this is the easier option) and proceed to check that they hold. The multivaluedness is in a constant which vanishes when you take derivatives.
  4. Sep 18, 2012 #3
    writing it in terms of r, theta and using CR, that shows it is analytic. how do i show it is harmonic? the e^iθ, after you take the ln, will no longer be harmonic!
  5. Sep 18, 2012 #4


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    What is the definition of a harmonic function? Harmonic and periodic are not the same thing.
  6. Sep 18, 2012 #5


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    Why do you think θ isn't harmonic? It's basically just arctan(y/x). That's perfectly fine harmonic function.
  7. Sep 19, 2012 #6
    sorry, perhaps I was confused, but I think the definition of a harmonic function should be one such that for f(x), ∂^2f/∂x^2 = kf(x). Or as wikipedia says:

    In mathematics, mathematical physics and the theory of stochastic processes, a harmonic function is a twice continuously differentiable function f : U → R (where U is an open subset of Rn) which satisfies Laplace's equation


    That's true for sine: ∂^2/∂x^2 (sin x) = -1*sin x
    Also true for cosine: ∂^2/∂x^2 (cos x) = -1*cos x

    However it is not true for arctan.

    for u = arctan(y/x), ∂^2u/∂x^2 =/= k*u. thus it is not a harmonic function.

    ∂^2u/∂x^2 (arctan(y/x) = -(2 x y)/(x^2+y^2)^2 instead.

    So that's why I'm confused. Yes, harmonic doesn't necessarily mean periodic, e^x is an example, but I'm having a hard time finding a function from ln(az) that can satisfy the 2nd derivative rule.
  8. Sep 19, 2012 #7


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    I think you need to look up Laplace's equation, because it is not ∂^2f/∂x^2 = kf(x).

    To quote Tai L Chow's Mathematical Methods For Physicists,

    So, when you are being asked to show that [itex]\ln(az)[/itex] is harmonic, you are really being asked to show that the two-variable functions describing its real and imaginary parts are harmonic. From the above quote, you should see that this is equivalent to showing that [itex]\ln(az)[/itex] is analytic over the specified region (which is where the Cauchy Riemann equations come in, as clamtrox suggested).

    Your textbook likely says something similar. Whenever you don't know how to tackle a problem, it is usually a good idea to open up your textbook and make sure that you understand the definitions of the relevant terms in the problem statement. Textbooks (good ones, anyways) are useful references to have.
    Last edited: Sep 19, 2012
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