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Suppose that $f=u+iv$ is holomorphic in an open set $\Omega$. Prove that $f$ is constant if $f$ is constant.

My attempt:

If $f\equiv 0$ then its trivial. If $|f|=c$ then $\partial_x|f|=\partial_y|f|=0$ so $\dfrac{u\partial_xu+v\partial_xv}{\sqrt{u^2+v^2}}=\dfrac{u\partial_yu+v\partial_yv}{\sqrt{u^2+v^2}}=0$ which implies that $\begin{cases}u\partial_xu+v\partial_xv=0\\ u\partial_yu+v\partial_yv=0 \end{cases}$ (since $\sqrt{u^2+v^2}=|f|$ is constant). But can't seem to be able to prove that $\partial_x u,\partial _x v=0$ (and thus prove $f'(z)=0$). What could I try next? Im kind of stuck.

Sebastián P. Pincheira
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2 Answers2

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If $w\in\Omega$ satisfied $f'(w)\neq0$,

then there would exist open sets $U\ni w$ and $V\ni f(w)$ such that the restriction of $f$ to $U$ is a bijection from $U$ to $V$.

https://people.ucsc.edu/~fmonard/Sp17_Math207/lecture7.pdf

In particular, $|f(z)|$ would not be constant, because as $v$ ranges over $V$, you obviously find that $|v|$ attains more than one value.

$\\$

So if you want $|f(z)|$ to be constant, this forces $f'(z)\equiv0$, implying that $f$ is constant.

Chris Sanders
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(1).

(1a).If $c=0$ then $f=|f|=0$ everywhere on $\Omega.$

(1b).If $c\ne 0$ then $(u,v)\ne (0,0)$. We have $c^2=u^2+v^2.$ So $0=\partial_x (c^2)=2u\partial_x u+2v\partial_x v$ and $0=\partial_y (c^2)=2u\partial_y u+2v\partial_y v.$ But $\partial_y u=-\partial_x v$ and $\partial_y v=\partial_x u$ because $f$ is holomorphic. Therefore $$2u\partial_x u+2v\partial_x v=0$$ $$-2u\partial_x v+2v\partial_x u=0.$$ This is not possible with $(u,v)\ne (0,0)$ unless $\partial_x u=\partial_x v=0,$ that is, unless $f'(z)=0.$

(1c).In either case we have $f'=0$ on $\Omega.$

(2).

Assuming dom $(f)=\Omega\ne\emptyset,$ choose $z_0\in\Omega$ and let $S=f^{-1}\{f(z_0)\}.$ For any $z\in S$ there is an open ball $B$ with $z\in B\subset\Omega,$ and by (1c) we have $f'=0,$ so $B\subset S.$ So $S$ is open in $\Bbb C,$ so $S$ is open in the subspace $\Omega.$ But $\{f(z_0)\}$ is closed in $\Bbb C$ and $f$ is continuous on $\Omega$ so $S$ is closed in $\Omega.$ Now $\Omega$ is connected and $S$ is a non-empty open-and-closed subset of the space $\Omega$, so $S=\Omega.$

DanielWainfleet
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