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$f$ is holomorphic in $\Omega$ such that $|f|$ is harmonic; we need to show that $f$ is constant.

Let

$$f=u(x,y)+iv(x,y)\Rightarrow |f|=\sqrt{u^2+v^2}\quad \rm{and}\quad \nabla^2|f|=0$$ right?

Also I have $u_x=v_y, v_x = -u_y$

$$\nabla^2 = {\partial^2\over \partial x^2}+{\partial^2\over \partial y^2},$$

so as $ \nabla^2|f|=0 $ we get

$$ u_{xx}+u_{yy}+v_{xx}+v_{yy}=0 $$

So now could any one show me how to proceed?

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    $\begingroup$ Try showing that $\Delta|f| = |f|^{-1}\left|\frac{\partial f}{\partial z}\right|^2$. The result follows quickly from there. $\endgroup$ Jun 20, 2013 at 5:48

1 Answer 1

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Try to complete the proof using the following hints:

1) If $f$ is nonzero in a neighborhood $U$ then $\log f$ exists.

2) $\log|f|$ is harmonic.

3) If $g$ is a harmonic function such that $\log|g|$ is also harmonic then $g$ is constant.

4) $f$ holomorphic and $|f|$ constant in a neighborhood implies $f$ is constant.

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