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Homework Help: Complex problem please help! f(z) = sqrt(|xy|) in x + iy form?

  1. Feb 1, 2009 #1
    1. The problem statement, all variables and given/known data

    In the title: f(z) = sqrt(|xy|)...show that this satisfies the Cauchy-Riemann equations at z=0, but is not differentiable there.

    2. Relevant equations

    Cauchy-Riemann just states that partial u partial x = partial v partial y and partial u partial y = - partial v partial x.

    3. The attempt at a solution

    I think all the partials du/dx, du/dy, dv/dx, and dv/dy are 0. Because f(0) is 0 in this case, right? Maybe that's wrong. Maybe I'm assumign too much by saying that if z=0, x and y also are 0.

    Anyway, then I just need to show the limit as h->0 for differentiation doesn't exist, right? If what I said above is right, than it would just be plugging in some x_0 and y_0 (if h = x_0 + iy_0) for x and y and than finding the limit for when x_0 = 0 and when y_0 = 0, and as long as those aren't equal then it's not differentiable.

    Does this sound right, or totally off?
  2. jcsd
  3. Feb 2, 2009 #2


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    You're supposed to take the partial derivatives first and then evaluate them at z=0; not the other way around!:smile:
  4. Feb 2, 2009 #3
    Wouldn't that mean the Cauchy-Riemann equations don't hold? I'm a little unsure on what u would be in this case.

    Do I need to separate sqrt(|xy|) into the real and imaginary parts? Or can I just assume all is real and then take the partial derivates of u, and the partials of v would just be 0, since there's no complex part?

    This is why I think Cauchy-Riemann wouldn't hold, but I'm not sure.

    Thanks so much for the response! I do appreciate it.
  5. Feb 2, 2009 #4
    Just to go ahead and try this: would partial u partial x be y(xy)^(-1/2)?
    And then partial u partial y be x(xy)^(-1/2)?
    And both partials of v be 0? This is assuming that sqrt(xy) is just the 'real' part...if f(z) takes the form u + iv.

    I have a feeling this is wrong, since Cauchy-Riemann is supposed to hold and doesn't hold using this method. But if it's wrong, how am I supposed to figure out what u and v are in cases like this?
  6. Feb 2, 2009 #5


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    Yes, x and y are real numbers, so |xy| is a positive, real number and so sqrt(|xy|) is real.

    Careful, what do you get for [tex]\frac{\partial u}{\partial x}[/tex] and [tex]\frac{\partial u}{\partial y}[/tex]....where do they equal zero?
  7. Feb 2, 2009 #6
    I'm not sure what you're saying here. I think you get that they don't exist or are undefined, because there would be 0 on the bottom of a fraction? But I'm not sure where you're going with this.
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