Show that the equation $e^z=az^n$ has $n$ solutions in the unit disc centred at the origin counting multiplicity.

Question

Let $a \in \mathbb{C}$ be such that $|a| > e$. Show that the equation $e^z=az^n$ has $n$ solutions in the disk $D_1(0)$ i.e. the unit disk centred at the origin counting multiplicity.

I think I'm supposed to use use Rouches theorem to solve this. Thanks in advance.

Answer 1

$|az^{n}| =|a| >e \geq |e^{z}|$ when $|z|=1$. Apply Rouche's Theorem to conclude that $e^{z}-az^{n}$ and $az^{n}$ have the same number of zeros in the open unit disk.