$e^z$ is conformal on $\CC$ but not injective

A conformal map need not be injective. The map \(f\colon\CC\to\CC\), \(f(z)=e^z\), is holomorphic with \(f'(z)=e^z\neq0\) for all \(z\), so it is conformal on \(\CC\). It is not injective because it is periodic: \(e^{z+2\pi i}=e^z\). It is also not surjective, since \(e^z\neq0\) for all \(z\in\CC\).