$f: \mathbb{C}\rightarrow\mathbb{C}$ Holomorphic and that $f(T)\in\mathbb{R}$

Question

Let $\mathbb{D}$ be the unit disk and $\mathbb{T}$ be the boundary of that unit disk.

(a)Show that f(0) is a real number.
(b) Show that for each $z\in\mathbb{D}$ we must have $f(z)\in\mathbb{R}$.
(c) Show that f must be a constant function.

For part (a) I have used the Cauchy integral formular $$f(0)=\frac{1}{2\pi i}\int\limits_{T}\frac{f(z)}{z}dz$$. Thus by using the parameterization $z=e^{i\theta}$ we obtain $f(0)$ equal to a real integral .

Also I know that for part (c) I can use the answer from part (b) together with the open mapping property of a holomorphic function.

But I would like a help to go with part (b)

Answer 1

Let$$g(w)=f\left(\frac{w+z}{1+\overline zw}\right).$$If $w\in\mathbb T$, then$$\left\lvert\frac{w+z}{1+\overline zw}\right\rvert=\left\lvert\frac{w+z}{\overline ww+\overline zw}\right\rvert=\left\lvert\frac{w+z}{\left(\overline w+\overline z\right)w}\right\rvert=1$$and therefore $g(w)\in\mathbb R$. But you have already proved that then $g(0)\in\mathbb R$. And this means that $f(z)\in\mathbb R$.

Answer 2

One way is to note that if $f=u+iv$ then $v$ is harmonic and so satisfies the maximum principle, which implies that $v=0$ everywhere in the disc so in fact $f$ send the disc into the reals, so it must be constant, by the open mapping theorem.