For $z = 1$ and $w = 0$ we obtain $f(0) = 0$. For $w = z \in \mathbb{C}^*$ 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 $|f(z)| = t_z \cdot 1 \cdot |z| \Leftrightarrow 1 = t_z$, for any $z \in \mathbb{C}^*$.

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.