let $\sum_{k=0}^{n}|f_{k}|^{2}=g$ then if g is constant then $\forall k\in[n].f_{k}$ is constant

Question

let $\sum_{k=0}^{n}|f_{k}|^{2}=g$ and then if g is constant then if $\left\{ f_{k}\right\} _{k=1}^{n}$ is holomorphic then : $\forall k\in[n].f_{k}$ is constant. My try: Each of the $f_k$ is has composition to 2 harmonic function $u+iv$ therefore its 2nd derivative is zero. therefore and because they have the same derivative they must be linear. but linear function accept constant in the entire plane iff they are constant function therefore $f_k must be constant.

Answer 1

Assume that $g$ is constant. Let $\langle \cdot, \cdot \rangle$ be the usual Hermitian form on $\mathbb C^n$. Take any $z_0$ in the domain. Let $f(z) = (f_1(z), \ldots, f_n(z))$. Then $z \mapsto \langle f(z), f(z_0) \rangle$ is holomorphic and by the Schwarz inequality:

$$ \lvert \langle f(z), f(z_0) \rangle \rvert \leq g$$

with equality at $z=z_0$. By the maximum principle $\langle f(z), f(z_0) \rangle = g$ on the connected component of the domain that contains $z_0$. And again by Schwarz (equality if and only if vectors are dependent) $f(z) = g(z) f(z_0)$ for some holomorphic $g$ with $\lvert g(z)\rvert = 1$. Then $g$ is constant and $f(z) = f(z_0)$. So $f$ is itself constant.