If $f$ is entire and $\mathrm{Arg}(f(z))=-\pi/2$ when $|z|=1$, then how to show that $f$ is constant using Liouville's theorem?

Question

If $f$ is entire and $\mbox{Arg}(f(z))=-\frac{\pi}{2}$ when $|z|=1$, then show that $f$ is constant.

All I need to prove is that f is bounded but I can't figure out how. I'd like someone's help.

Answer 1

The harmonic function $u(z)=\textrm{Re}f(z)$ satisfies $u=0$ on $|z|=1$, so $u\equiv 0$ by the uniqueness of solutions to the Dirichlet problem $\Delta u=0$ on $D$, $u=0$ on $\partial D$.

Answer 2

Let $g\colon \mathbf C\setminus \{-i\} \rightarrow \mathbf C\setminus\{1\}$ be a Moebius transformation mapping the real line $\mathbf R$ onto the unit circle minus the point $1$. Let $h=i\cdot f\circ g$. Then $h$ takes real values on $\mathbf R$. It follows that $h(\bar z)=\overline{h(z)}$ for all $z\in\mathbf C\setminus\{\pm i\}$. Since $h$ can be extended over $i$, it can then also over $-i$. Hence $h$ is entire. Since $g(z)\rightarrow1$ if $z\rightarrow\infty$, and $f$ is holomorphic at $1$, the entire function $h$ has a limit as $z$ tends to infinity. In particular, $h$ is bounded, and therefore constant by Liouville's theorem.

Answer 3

Suppose that $f(z)=u(z)+iv(z)$. Consider the functions $$g(z) = e^{f(z)},~ ~~~ h(z) = e^{-f(z)}.$$ Note that, since $u|_{|z|=1} \equiv 0$, it follows that $$|g(z)|=|h(z)| = 1$$ on $|z|=1.$ Thus, by Maximum Modulus Principle, we have, $|g(z)| \leq 1$ and $|h(z)| \leq 1$ on $|z| \leq 1$. However,

$$|g(z)|=|e^{u(z)}|$$ and $$|h(z)| = |e^{-u(z)}|$$

which implies, $$|e^{u(z)}| \leq 1,~~~|e^{-u(z)}| \leq 1$$ on $|z| \leq 1.$ Hence, at least one of $g$ or $h$ is constant, and therefore $f$ is constant.

$\textbf{NOTE :}$ It is clear that this argument will hold even under the assumption that $f$ maps $\mathbb{S^{1}}$ to a line, not necessarily the imaginary axis. Simply apply translations and rotations accordingly.