1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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

    Dick

    User Avatar
    Science Advisor
    Homework Helper

    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

    Dick

    User Avatar
    Science Advisor
    Homework Helper

    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

    Dick

    User Avatar
    Science Advisor
    Homework Helper

    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.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Show a limit doesn't exists, complex case
Loading...