$f(z_1 z_2) = f(z_1) f(z_2)$ for $z_1,z_2\in \mathbb{C}$ 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.
 A: 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,
A: Let $k=f'(1)$.  Define $f_a(z)=f(az)=f(a)f(z)$.  Differentiating by $z$ we get:
$$f_a'(z) = a f'(az) = f(a) f'(z)$$
Setting $z=1$, we get that $a f'(a) = k f(a)$ for arbitrary $a$.  Writing this as an equality of power series and we get, that $k$ must be a non-negative integer, and that $f(a)=a^k$, or that $f(a)=0$ for all $a$.
[This assumes that $f$ is defined on $1$ and $0$.  Obviously, there are examples where it isn't defined on $z=0$ - the cases where $k<0.$]
If $f(1)$ is defined, then we don't need $f(0)$ because we can rewrite the above as: $f'(a)/f(a) = k/a$ when $a\neq 0$ and $f(a)\neq 0$.  But the left side is the derivative of $\log f(z)$ for some branch of the natural log, and the right side is the derivative of $k \log z$ for some branch of $\log$. So $f(z) = e^{\log f(z)} = K e^{k \log z} = K z^k$ in some region around $a$, and hence all of the domain of $f$ (assuming the domain is connected?)
If $f(a)=0$, on the other hand, then $f(ax)=0$ for all $x$ near $1$, which would mean that $f$ is zero on some ball around $a$, so $f$ is identically zero.
