$z \rightarrow z^n$ is unbranched on $\mathbb{C}^*$

Question

Suppose $p: \mathbb{C} \rightarrow \mathbb{C}$ is a non-constant holomorphic map. A point $y \in \mathbb{C}$ is called a branch point of $p$, if there is no neighborhood $V$ of $y$ such that $p|_V$ is injective. The map $p$ is called an unbranched holomorphic map if it has no branch points.

Now, a basic exercise following this definition was to prove that the only branch point of the mapping $p: \mathbb{C} \rightarrow \mathbb{C}$ given by $p(z) = z^n (n \geq 2)$, is $0$.

I can see that $0$ is a branch point because - if I take any neighborhood $V$ of $0$, then $\exists 2\epsilon > 0$ such that $D(0, 2\epsilon) \subset V$. Hence $$p\left(\epsilon \exp\left(\frac{2\pi \iota}{n}\right)\right) = p(\epsilon) = \epsilon^n $$

I am stuck with the next part, which is, I am not able to construct an explicit neighborhood $V$ of $z \neq 0$ such that $p|_V$ is injective.

Any help will be appreciated. Thanks.

Answer 1

If $z_0\ne0$, then $p'(z_0)=n\,z_0^{n-1}\ne0$. The inverse function theorem implies that $p$ is injective on a neighborhood of $z_0$.

If you do not know the inverse function theorem, you can argue as follows. Let $z_0=r_0e^{i\theta_0}$ with $r>0$. Choose a neighborhood of $z_0$ of the form $$ U_\alpha=\{re^{i\theta}:r>0,\ \theta_0-\alpha<\theta<\theta_0+\alpha\} $$ with $\alpha>0$ small enough, and show that the equation $z^n=w$ has a unique solution $w\in U_\alpha$.

Answer 2

Say that $p$ is a branch point. Therefore, therefore, we can find $\varepsilon>0$ such that for $z-p$ we can find $f(p)=f(z)$.

Thus:

$$ p^n-z^n=(p-z)(p^{n-1}+p^{n-1}z+...+pz^{n-2}+z^{n-1})=0 $$

Then either $p=z$, which does not prevent injectivity or $p=-z e^{ki\pi/n}$. But in this case:

$$ |p-z|=|p||1-e^{\theta i}|=|p|(2-2\cos\theta)<\varepsilon $$ Then either $|p|<2\varepsilon$ which implies $p=0$ or $\cos\theta=1$, which implies $z=p$, and that does not conflict with injectivity.

Therefore $p=0$.