75th Romanian Mathematical Olympiad

For $z=1$ and $w=0$ we obtain $f(0)=0$. For $w=z\in {\mathbb{C}}^{\ast }$ we obtain $|f(z)|=|z|$; this equality is also verified for $z=0$, so $|f(z)|=|z|$, for any $z\in \mathbb{C}$. We have $|f(1)|=1$ and, for $w=1$, we obtain $2|z|=|f(z)+zf(1)|\le |f(z)|+|zf(1)|=2|z|$ for any $z\in \mathbb{C}$. It follows that we have equality in the triangle inequality. Therefore, for every $z\in \mathbb{C}$, there exists $t_z\in \mathbb{R}$, $t_z\ge 0$, such that $f(z)=t_z\cdot f(1)\cdot z$. We apply the modulus to both members and, after simplification, we have $|f(z)|=t_z\cdot 1\cdot |z|\iff 1=t_z$, for any $z\in {\mathbb{C}}^{\ast }$.

We can conclude that $f(z)=cz$ for any $z\in \mathbb{C}$, where $c=f(1)$ is a complex number with modulus equal to 1. It is immediately verified that any function of the form $f(z)=cz$, where $c\in \mathbb{C}$, $|c|=1$, is a solution of the functional equation in the statement.