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Complex Differentiable

  1. May 24, 2012 #1
    f'(z0)=[itex]\stackrel{lim}{x\rightarrow0}[/itex] [itex]\frac{f(z0-z)-f(z0)}{z}[/itex]

    Hi, I'm attempting to use the above equation to show where z0 is not differentiable at some point z0 for the equation

    f(z) = |z|2

    I was wondering how I could go about doing this?
    I tried letting z0 = a + bi, and z = x + yi and got [itex]\stackrel{lim}{x\rightarrow0,y\rightarrow0}[/itex] [itex]\frac{2ax+2yb-2ab+x^2+y^2}{x+yi}[/itex]

    I also tired the Cauchy Riemann Equations and let u(x,y) = x^2 + y^2 v(x,y) = 0
    f(x,y) = u(x,y) + iv(x,y)
    |x+yi| = [itex]\sqrt{x^2+y^2}[/itex]
    So 2x = 0, and -2y = 0.
    So the equation is differentiable when (x,y) = (0,0)?
    Any help is appreciated.
  2. jcsd
  3. May 24, 2012 #2
    One thing about the Cauchy Riemann equations is that they easily imply that u and v are each harmonic functions...
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