Is $f(z) = z^{2}$ an automorphism of $\mathbb{C}$?

Question

Let $f: \mathbb{C} \rightarrow \mathbb{C}$ where $f(z) = z^{2}$. I claim $f$ is an automorphism of $\mathbb{C}$ as a vector space. First $\operatorname{ker}(f) = 0$ and since $\mathbb{C}$ is algebraically closed then for all $z$ in $\mathbb{C}$, $\sqrt{z}$ exists and so $f$ is surjective.

Answer 1

We have $f(-z)=f(z) \ne -f(z)$ for $z \ne 0$. This shows that $f$ is neither injective nor linear.