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I want to find something like $f(z) = |z|$ (complex function that get's only real values) the problem here is that $f(z)$ is not analytic.

d_e
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2 Answers2

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If $f=u+iv$ with $v=0$, by C-R equations: $$u_x=v_y=0,$$ $$u_y=-v_x=0,$$ So $u$ is...

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By the open image theorem, nonconstant analytic functions have an open image, so if $f$ is analytic and only takes real values, its image is included in $\mathbb{R}$, so can't be an open set in $\mathbb{C}$. Therefore $f$ is constant.

Cauchy-Riemann equations can also lead to the same conclusion.