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!

Complex show differentiable only at z=0

  1. Feb 1, 2009 #1
    1. The problem statement, all variables and given/known data
    Show that f(z) = zRez is differentiable only at z=0,
    find f'(0)

    3. The attempt at a solution

    This should be easy. I find the limit as z_0 approaches 0 of [f(z+z_0) - f(z)]/(z_0) for this function...expand it out, simplify, and find what the limit is when z_0 is purely imaginary vs purely real. I did this, and I got x for the real part and x_0 + 2x + iy for the imaginary part. They're different, so it's not differentiable everywhere.

    But apparently it's differentiable only at z=0. So I tried plugging in z=0 and resolving, and I got different values for z_0 real and imaginary.

    I get, for the limit when z_0 is only real: 1
    When z_0 only imaginary: x_0.

    What's going on? Is there some other way to differentiate a complex function that's not using the limit?
  2. jcsd
  3. Feb 2, 2009 #2


    User Avatar
    Homework Helper
    Gold Member

    Try your expansion and simplification again. When [itex]z_0[/itex] is real you should find that [tex]\lim_{z_0\to 0} \frac{f(z+z_0) - f(z)}{z_0}=2x+iy[/tex] and when [itex]z_0[/itex] is imaginary you should find that [tex]\lim_{z_0\to 0} \frac{f(z+z_0) - f(z)}{z_0}=x[/tex]
    Last edited: Feb 2, 2009
  4. Feb 2, 2009 #3
    I mixed up my original message. I actually got x for the imaginary part and x_0 + 2x + iy for the real part. I still don't see how the x_0 was eliminated in your version of the expansion, above.

    But the main issue is the fact that I get different values in the second part, evaluating the limit at z=0. I get different values for the real and imaginary parts, which means the limit doesn't exist. But according to the problem it should exist! I must just be doing the math wrong.

    Am I at least on the right track with the method of finding the limit as z_0 goes to 0, setting z=0, and solving it all out and seeing if the real and imaginary parts of z_0 both go to the same value?

    Thanks again for the help!
  5. Feb 2, 2009 #4


    User Avatar
    Homework Helper
    Gold Member

    Doesn't x_0 go to zero when z_0 goes to zero?:wink:

    You are doing the math wrong!:smile:

    At z=0, x and y are both zero too aren't they?....Last time I checked 0=0 was a true statement:wink:
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?

Similar Discussions: Complex show differentiable only at z=0