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Math Help - prove the the conjugate of a holomorphic function f is differentiable iff f'=0

  1. #1
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    prove the the conjugate of a holomorphic function f is differentiable iff f'=0

    This probably has a simple solution, but I cannot find it so far...

    Quote Originally Posted by problem statement
    Suppose the complex-valued function f is holomorphic about zero. Show that g:=\overline{f}, i.e. the conjugate of f, is differentiable at z if and only if f'(z)=0.
    The first direction follows fairly directly from the Cauchy-Riemann equations. However, I still need to show that f'(z)=0 implies g'(z) exists, and I can't seem to make it happen. Any help would be much appreciated.

    Thanks!
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  2. #2
    MHF Contributor FernandoRevilla's Avatar
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    If f=u+iv and z_0=x_0+iy_ use:


    f'(z_0)=\dfrac{\partial u}{\partial x} (x_0,y_0) +i\dfrac{\partial v}{\partial x}(x_0,y_0)


    Fernando Revilla
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  3. #3
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    Hmm. I guess I don't see how that helps us here.
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