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Homework Help: Show a limit doesn't exists, complex case

  1. Oct 4, 2011 #1
    1. The problem statement, all variables and given/known data

    [itex]f(z) = |z|[/itex]
    I'm looking to show that [itex]f'(z)[/itex] does not exist for any [itex]z \in ℂ[/itex].

    2. Relevant equations

    [itex]f'(z) = \lim_{z_0 → 0}{\frac{f(z) - f(z_0)}{z - z_0}} [/itex]

    [itex]z = x + iy = Re(z) + i Im(z)[/itex]

    [itex]|z| = \sqrt{x^2 + y^2}[/itex]

    3. The attempt at a solution
    Clearly I just have to show that [itex]\lim_{z_0 → 0}{\frac{f(z) - f(z_0)}{z - z_0}}[/itex] does not exist. However, I'm confused about how to do this. I'm unsure of how to show that a limit doesn't exist in complex analysis.

    I tried to take the limit in two cases, when [itex]Re(z) = Re(z_0)[/itex] and then again when [itex]Im(z) = Im(z_0)[/itex]. Is this the correct approach?
  2. jcsd
  3. Oct 4, 2011 #2


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    Have you done the Cauchy-Riemann equations?
  4. Oct 4, 2011 #3
    I was only trying to prove it with limits because it's an exercise in my complex analysis book that comes before the Cauchy-Riemann equations are introduced.

    [itex]f(z) = U(x,y) + iV(x,y)[/itex]

    [itex]f(z) = |z| = |x + iy| = \sqrt{x^2 + y^2}[/itex]

    [itex]U(x,y) = \sqrt{x^2 + y^2}[/itex]
    [itex]V(x,y) = 0[/itex]

    [itex]U_x = 0.5(x^2 + y^2)^{-1/2}(2x) = x(x^2 + y^2)^{-1/2}[/itex]
    [itex]U_y = y(x^2 + y^2)^{-1/2}[/itex]
    [itex]V_x = 0[/itex]
    [itex]V_y = 0[/itex]

    [itex]U_x ≠ V_y[/itex]
    [itex]U_y ≠ -V_x[/itex]

    However, does the derivative exist when

    [itex]x(x^2 + y^2)^{-1/2} = 0[/itex]
    [itex]y(x^2 + y^2)^{-1/2} = 0[/itex]

    i.e. when z = 0?
    Last edited: Oct 4, 2011
  5. Oct 4, 2011 #4


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    U_x=x/sqrt(x^2+y^2). That's not what you wrote. And the derivatives don't exist at (x,y)=(0,0) either, for the same reason the derivative of |x| with respect to x doesn't exist at x=0.
  6. Oct 4, 2011 #5
    Yeah, I forgot the (-1/2) exponents when I typed it out.

    Okay, so z = 0 is a "potential" solution after I check the C-R equations, but then when I check it case by case, I see that the derivative doesn't exist when z = 0. Is this a correct?
  7. Oct 4, 2011 #6


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    Right. U_x and U_y at z=0 have the form 0/0. That doesn't tell you much. But if you look at the details, they don't exist at z=0 either.
  8. Oct 4, 2011 #7
    Thank you! I'm still confused about limits in the complex plane and my original approach to this problem, but I'll save that for another time and another problem :) I really appreciate your help.
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