Prove that there does not exist a function $f: C\rightarrow C$, excluding the origin, with the following two properties:

$f(zw)=f(z)f(w)$ for all $z,w$ $\in C$ and $(f(z))^2=z$ for all $z$ $\in C$

My attempt:

Assume on the contrary that such a function exists. The existence of this function leads to the implication that 

$f(zw)=f(z)f(w)$ implies $(f(z))^2=f(z)f(z)=f(zz)=f(z^2)=z$

I consider the identity map such that
$f(zw)=zw$ and show that the implication does not hold for all $z$. My question is whether the the counterexample using the identity map is valid.