functional equations - $f(z_1 z_2) = f(z_1) f(z_2)$ for $z_1,z_2in mathbb{C}$ then $f(z) = z^k$ for some $k$

Same as my previous question except domain is complex. I tried assuming that the function was analytic, so for $z_1=z_2=z$ , $f(z^2) = f(z)^2$ $$\sum_{n=0}^\infty a_n z^{2n}=\left(\sum_{n=0}^\infty a_n z^n\right)^2$$ and try to solve it. Ideally, I should have something like $a_n=0$ for all but one k. and that should prove it.

This question, with $z_1=z_2$ was asked today but I have not been able to understand their answers.

Answer

This question is very similar to the other.

Notice that the constant zero function is a solution. For the remainder of this answer, I assume that $f$ is not identically $0$.

First, let $z_1=0$. Then $f(0)=f(0)f(z_2)$ for all $z_2$. Lets split into cases since either $f(0)=0$ or $f(0)\neq 0$.

Case 1: Suppose $f(0)\neq 0$. Then in the equation $f(0)=f(0)f(z_2)$ we can divide through by $f(0)$ and conclude that $f(z_2)=1$ for all $z_2$, so $f$ must be the constant function $1$.

Case 2: Suppose $f(0)=0$. As $f$ is analytic, we may write $f(z)=z^k g(z)$ for some analytic $g$ where $g(0)\neq 0$. This function $g$ must satisfy the same functional equation as $f$ (why?). Hence by case 1, it follows that $g=1$, and we conclude $f(z)=z^k$.

Hope that helps,