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Let be $f$ a complex function in zone $\Omega$ can someone help me how to prove that if $f$ and $f^2$ are harmonic in $\Omega$ then $f$ and $f^2$ are holomorphic in $\Omega$

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What do you mean by harmonic here? Usually harmonic functions are assumed to be real valued. –  mrf Jun 28 '12 at 19:20
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@mrf but the notion makes just as much sense for complex and vector-valued functions, does not it? –  user31373 Jun 28 '12 at 19:39
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@Adi Dani they can be antiholomorphic. –  Andrew Jun 28 '12 at 19:47
    
@LeonidKovalev it does, but since it's a bit non-standard, I wanted to make sure (or at least show the OP that it's non-standard) –  mrf Jun 28 '12 at 19:52
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up vote 2 down vote accepted

$f$ is harmonic if $f_{z\bar z}=0$ (you should check this if you haven't seen this before). Another way to phrase this: $f$ is harmonic if and only if $f_z$ is holomorphic. We are told that $f_z$ and $(f^2)_z$ are holomorphic. By the chain rule, $(f^2)_z=2ff_z$, and since $f_z$ is holomorphic, you should be able to conclude that $f$ is holomorphic as well.

[Added] Or antiholomorphic, in case $f_z\equiv 0$. Thanks, @Andrew.

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