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I want to prove that if $f$ is a non-constant holomorphic function and has no zero points in an open domain $D$, then $|f|^p(p>0)$ is not harmonic in $D$.

I tried to use $\Delta=4 \frac{\partial^{2}}{\partial \bar{z} \partial z}$ but I found the calculation is too long. Is there any similar solution like this post?

Yuri
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    Show for $|f|^2$ first as that is $f\bar f$ so easy to apply laplacian; then since $f$ has no zeroes, it is locally $g^{2/p}$ for some holomorphic $g$ and apply above – Conrad May 17 '22 at 15:55
  • @Conrad : Thank you for the comment. Would you mind explaining "it is locally $g^{2/p}$ for some holomorphic $g$" in detail? I didn't really understand the statement. – Yuri May 18 '22 at 13:00
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    $f$ has no zeroes, so in a neighborhood of any $w \in D$ it has a holomorphic logarithm $h_w$ (it may not have a global logarithm as $1/z$ in the punctured unit disc shows), and then with $g_w=\exp (ph_w/2)$ one has $g_w^{2/p}=f$ so $f^p=g_w^2$ – Conrad May 18 '22 at 14:46
  • I see, thank you very much! – Yuri May 18 '22 at 15:38

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