Htykuut

Let $f in mathbb{C}[z]$ a complex polynomial such that $f(mathbb{D}) subset mathbb{R},$ where
$$
mathbb{D} = { z in mathbb{C} : |z| = 1 }
$$
Show that $f$ is constant.

My attempt is: define a function $g(z) = e^{if(z)}$ and plug some roots of unity in $g,$ but how can I conclude from this?
I know that if $Omega$ is open, connected and $f(Omega) subset mathbb{R},$ then $f$ is constant.

Let $B(0,1)$ be the unit open ball centered at the origin, and let $D^2 = B(0,1) cup mathbb{D}$ be the closed unit ball centered at the origin.

If $f$ is not constant, then by the open mapping theorem, $f(B(0,1)) = Omega$ is an open subset of $mathbb{C}$. Since $mathbb{D}$ is the boundary of $B(0,1)$ it maps into the boundary of $Omega$. Since $D^2$ is compact, $f(D^2)$ is compact; so $f(D^2)$ is closed and bounded.

So, $mathbb{D}$ maps onto the boundary of $Omega$. But this can only happen if $f(B(0,1)) = mathbb{C} - f(mathbb{D})$. This is indeed open because $f(mathbb{D})$ is a compact set and hence closed and bounded. However, this means that $f(B(0,1))$ is not bounded, which is a contradiction because $f(D^2)$ was deduced to be bounded, and $f(D^2) supset f(B(0,1))$.

Thus, we have a contradiction. Hence, $f$ must be constant.

By the way, your notation is a bit odd because I have seen $mathbb{D}$ to be usually used to denote the closed unit disc, rather than the circle. Perhaps it would be appropriate to double check the source of the question.

If we set
$$
u(x,y)=mathrm{Im},f(x+iy),
$$
then $u$ is a harmonic function, which vanishes on the unit circle. Maximum (together with Minimum) Principle for harmonic function imply that
$$
max_{x^2+y^2le 1}{u(x,y)}=max_{x^2+y^2=1}{u(x,y)}=0
quadtext{and}quad
min_{x^2+y^2le 1}{u(x,y)}=min_{x^2+y^2=1}{u(x,y)}=0
$$
Hence, $uequiv 0$ in the closed unit disk, and hence $uequiv 0$ everywhere. (Identity Principle.)

Thus $f$ takes only real values, and hence $f$ is constant.

I think we can do this from scratch, especially since we are given that $f$ is a polynomial:

Let $mathbb T$ be the unit circle, and $f$ a polynomial of degree $n$.

For $z=e^{it}$, consider the real number

$f(e^{it})=a_0 +a_1e^{it}+a_2e^{2it}+cdots a_{n}e^{nit}.$

The imaginary part of this must be equal to zero, so

$Im a_0+sum^{n}_{k=1}(Re a_ksin kt+Im a_kcos kt)=0$.

But now, linear independence of the set $left {1, sin kt,cos kt right }^{n}_{k=1}$ on $mathbb T$ implies that

$Im a_0=a_1=cdots=a_n=0$ so $f(z)=Re a_0$.

The imaginary part $text{Im};f(z)$ of $f(z)$ is harmonic, and zero on the unit circle, hence by the maximum principle for harmonic functions, we conclude that $text{Im};f(z)$ is identically zero in the ball. Thus $f$ is real in the ball. Of course a non-constant real function cannot have complex derivative at $0$; in particular $f$ is not a polynomial.