# Let $f$ be an analytic function such that if $|z|=\frac{1}{2}$ then $f(z)\in \mathbb{R}$. Prove that $f$ is constant.

Let $f: \mathbb{D} \rightarrow \mathbb{C}$ be an analytic function such that if $|z|=\frac{1}{2}$ then $f(z)\in \mathbb{R}$. Prove that $f$ is constant. ($\mathbb{D}$ is the unit disk)

Any hints are appreciated

• An idea I haven't thought through at all: Conformally map the disk to the upper half plane, and then translate so that the given circle is sent to the real line. Then look at the Cauchy-Riemann equations. – user61527 Dec 17 '13 at 3:01
• @T.Bongers That's interesting !, but isn't that result for "a particular" analytic map (say, a linear fractional transformation) ? – the8thone Dec 17 '13 at 3:57
• You'd compose the function $f$ with a linear fractional transformation, and prove that the composition is constant by C-R. – user61527 Dec 17 '13 at 3:58

Hint

$f(z)=u(z)+iv(z)$ (say)

Define $g(z)=e^{i f(z)},h(z)=e^{-i f(z)}$ and apply Maximum Modulas Principle to them.

Then what can you say about $v(z)$ on the disk $\{z:|z|\le {1\over 2}\}$?

• Thank you so much ! I see that on the circle with radius $\frac{1}{2}$, the analytic function $g(z)$ attains the absolute value 1. If I know (Is this clear ???) that the function $g(z)$ has absolute value 1 at some point inside the circle of radius $\frac{1}{2}$, I can conclude that $g(z)$ is a contant and therefore $f(z)$ is a constant. – the8thone Dec 17 '13 at 3:18
• Thanks ! But still, that is not clear to me why the function g has abs value one inside that circle :-| – the8thone Dec 17 '13 at 3:22
• Maximum Modulas Principle forces that ..just have a look at the maximim modulas principle statement. – Marso Dec 17 '13 at 3:24
• I just looked at the Max.Mod.Principle again : Let f be a function holomorphic on some connected open subset D of the complex plane C and taking complex values. If $z_0$ is a point in D such that $|f(z_0)|\geq |f(z)|$ for all z in a neighborhood of $z_0$, then the function f is constant on D. Now my question is : if $|z_0|=1/2$ I am sure that $|f(z_0)|\geq |f(z)|$ for all z "iniside" the circle of radius 1/2 (by analyticity of g) not necessarily for points z in a neighborhood of $z_0$ that fall outside the circle of radius 1/2. I don't know what I am missing :-( – the8thone Dec 17 '13 at 3:38
• @Roozbeh-unity : I am not sure it is the right way to show that $|g(z)|=1$ for some $|z|<1/2$ (or maybe I misunderstand the hint). However I suggest you to consider the maximum AND the minimum of $|g|$ and note that $g$ doesn't vanish. – user10676 Dec 17 '13 at 17:50