Let $f$ be an entire function and suppose $|f(z)| = |z|$ for $|z| < 1$. Prove that $f(z) = λz$ for some $λ ∈ \mathbb{C}$ with $|λ| = 1$.
My idea is to show that the function $g$, defined as $g(x)=\begin{cases} \frac{f(x)}{x}& x\neq0,\\ 1&x=0\end{cases}$ is constant, using Liouville's theorem, and then show that the modulus of that constant is equal to $1$. But I am having trouble proving that $g$ is bounded and entire. Any ideas of how to do this?
Let's use nothing special.
With $f$ entire, suppose $U\subset\Bbb C$ is open and $|f(z)|=|z|$ for all $z\in U$. For $0\ne z_0\in U$, write $$ f(z_0+h)=uz_0+\lambda h + h^2g(h)$$ where $u=\frac{f(z_0)}{z_0}\in S^1$, $\lambda =f'(z_0)$, and $g$ is continuous and therefore bounded in a neighbourhood of $z_0$. Let $h=tz_0$ with $t\in(-1,1)$. Then $$\begin{align}(1+t)|z_0|&=|z_0+h| \\&=|f(z_0+h)|\\&=|uz_0+\lambda tz_0+t^2z_0^2g(h)|\\&\approx |u_0+\lambda t||z_0|\\&=\left|1+\frac\lambda ut\right||z_0|\end{align}$$ (where "$\approx$" means "up to $O(t^2)$"). This is possible only when $\lambda=u$.
This means $f(z)=zf'(z)$ for all $z\in U$, so $$(\ln f(z))'=\frac{f'(z)}{f(z)}=\frac 1z=(\ln z)'$$ $$\ln f(z)=\ln z+\textit{const} $$ $$\tag1 f(z)= z\cdot\textit{const} $$ and the last constant is clearly of modulus $1$.
As $f$ is entire, $(1)$ must hold throughout.
$|f(z)| = |z|$ implies $f(0)=0$ and so we can write $f(z)=zg(z)$ with $g$ entire. Moreover, $|g(z)|=1$ inside the unit disk. Therefore (*), $g$ is constant in the unit disk. Since $g$ is entire, it is constant everywhere.
(*) A holomorphic function with constant magnitude must be constant.
