Analytic bijective function is either $az$ or $\frac{a}{z}$ I am trying to solve the following problem:

Let $\mathbb{C}^* = \{z: 0 < |z| < \infty\}$ and $f: \mathbb{C}^* \to \mathbb{C}^*$, analytic and bijective function. Show that $f(z) = az$ or $f(z) = \frac{a}{z}$ for some $a \in \mathbb{C}$

I don't know where to start and would appreciate a hint.
Thanks!
Edit: I'm adding my (fairly) detailed solution for anyone who needs it. If you are looking for hints, you will find them in the comments.
 A: My solution:


*

*0 is not an essential singularity of $f$. To show this, assume that it is. Choose some $w,z \in \mathbb{C}^*$ such that $f(z) = w$. From the local mapping theorem, we have small neighborhoods of w and z - $U_w, U_z$ respectively, such that for every $w' \in U_w$, there's $z' \in U_z$ such that $f(z')  =w'$. Now take a small neighborhood of 0 that does not intersect with $U_z$ - $U_0$. From Casorati–Weierstrass, $f(U_0)$ is dense in $\mathbb{C}$ and particularly in $U_w$ hence $f$ is not bijective, which is a contradiction.

*If 0 is a removable singularity of $f$, then we continue it naturally to $F$ over the entire plane. $F(0) = 0$ - we show this using a similar method to what we did in #1.

*It is a 0 of order 1, otherwise, using the Local Mapping Theorem, $F$ is not bijective in a small neighborhood of 0.

*$F$ is a nonconstant entire function that doesn't have an essential singularity at $\infty$ (because it is bijective), hence it is a polynomial. We can write: $F(z) = az(\lambda_1 - z)...(\lambda_n - z)$ where the lambdas are the function's other roots aside from 0, but we know that $f \neq 0$ so $F = az$.

*If 0 is a pole of $f$, define $g(z) = \frac{1}{f}$. From here we continue like before to show that $g = az$, hence $f = \frac{1}{az}$, which is what we wanted to show.

