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Let $f: G\rightarrow \mathbb{C}$ be a holomorphic function on a domain. Let $\left[\Re{(f)}\right]⁴+\left[\Im{(f)}\right]⁴$ have a local maximum in $G$. Why is $f$ than already constant?

If I could show that the absolute of $f$ had a maximum I'd be done. Also, I could maybe use the Cauchy-Rieman differential equations but that seems very inelegant...

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It is basically the same as a local maximum of $\lvert f\rvert^2$. A non-constant holomorphic function on a domain is an open mapping, so for every point $z_0 \in G$ and every open neighbourhood $U$ of $z_0$, $f(U)$ is an open neighbourhood of $f(z_0)$, so there is a point $z_1 \in U$ with $f(z_1) = c\cdot f(z_0)$ for some $c > 1$ (if $f(z_0) \neq 0$; the case $f(z_0) = 0$ is clear) and that implies

$$[\Re f(z_1)]^4 + [\Im f(z_1)]^4 > [\Re f(z_0)]^4 + [\Im f(z_0)]^4,$$

so $[\Re f]^4 + [\Im f]^4$ does not have a local maximum in $z_0$.

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Functions $|\Re(f)|^4$ and $|\Im(f)|^4$ are subharmonic, and so is their sum. By the maximum principle, the maximum of a subharmonic function $|\Re(f)|^4+|\Im(f)|^4$ cannot be achieved in the interior of domain $G$ unless the function is constant. Hence  $|\Re(f)|^4+|\Im(f)|^4=C>0$  on  $G$  for a nonzero holomorphic $f$, which implies that $f$ is constant on  $G$.

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Assume that this local maximum is achieved at $z_0=x_0+iy_0$, where $f(z_0)=u_0+iv_0$.

If $f$ in non-constant, then it satisfies the open mapping theorem, i.e., the image of $G$ is an open subset $U$ of $\mathbb C$, with $u_0+iv_0\in U$. This means that there is such $\varepsilon>0$, such that $$ B_\varepsilon(u_0+iv_0)\subset U. $$ Say for example that $u_0,v_0\ge 0$. Then $u_0+\varepsilon/2+i(v_0+\varepsilon/2)\in B_\varepsilon(u_0+iv_0)\subset U$, and $$ (u_0+\varepsilon/2)^4+(v_0+\varepsilon/2)^4>u_0^4+v_0^4, $$ which is a contradiction.

Thus $f$ is constant.

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