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Let $G\subset \mathbb{C}$ be a region, $f$ a holomorphic function.

Then it does hold that:

  1. If $f(G) \subset \mathbb{R} \Rightarrow f$ is constant

  2. If $|f(z)|=1$ for all $z \in G$, then $f$ is constant.

Proof of 1:

$f$ is holomorphic, so it holds that $$u_x=v_y; u_y=-v_x$$ where $f(z) = u(x,y)+iv(x,y)$, then $u_x=0$ and $u_y=0$ so $u,v$ both must be constant

Proof of 2:

$$|f(z)|= 1 \Rightarrow u^2+v^2=1 $$

taking the partial derivatives:

$$2uu_x+2vv_x=0 , 2uu_y+2vv_y= 0$$

putting inside the Cauchy-Riemann equations:

$$uu_x-vu_y=(u^2u_x^2-2uvu_xu_y+v^2u_y^2)=0; uu_y+vu_x=(u^2u_y^2+2uvu_xu_y+v^2v_y^2)=0$$

so either $u=v = 0$ or $u_x=u_y=0$ so $$\Leftrightarrow (u^2+v^2)(u_x^2+u_y^2)=0.$$

If everything was done correctly so far, wouldn't switching $1$ with a constant $M$ give a proof for Liouville's theorem?

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Liouville's theorem only requires a bounded norm, not constant, so your proof doesn't prove Liouville's theorem. –  Ragib Zaman Sep 8 '12 at 11:24
The Liouville's theorem that I know of is about entire functions, and you can't really hope for it to be true for functions of bounded domain. Also, partial derivative of $1$ is $0$. –  tomasz Sep 8 '12 at 11:48

1 Answer 1

Both claims follow at once from the fact that a non-constant holomorphic function is an open map: Apparently, the image of $f$ is not open in $\mathbb C$.

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