I'm working on the following problem:
Let $f : U \rightarrow \mathbb{C}$ be holomorphic on an open set $U$ and suppose that $f$ has a zero of order $n \geq 2$ at $0$. Show that there is a small disc around $f(0)$ such that the inverse image $f^{−1}(\mathbb{R} \cap f(U))$ consists of $n$ disctinct arcs through the origin that are equally spaced.
To get started I have that a zero of order $n$ means I can factor $f$ and locally around zero, $f(z)=z^ng(z)$ with $g(z) \neq 0$ so that $g(z)$ is some complex number say $re^{i\theta}$. Then in a small neighborhood around zero, $f(z)^{\frac{1}{n}}=zg(z)^{\frac{1}{n}}=zr^{1/n}e^{i\theta/n}$. I'm having trouble arguing that there will be multiple arcs in the preimage mapping to $\mathbb{R} \cap f(U)$ - for this, wouldn't I want an expression like $e^{i\theta n}$?