Conformal map that maps $\Bbb C−\{0\}$ onto itself is of the form $az$ for some $a ∈ \Bbb C−\{0\}$, or $zb$ for some $b ∈ \Bbb C − \{0\}$

Question

Hint: Note that $0$ and $∞$ are isolated singularities of $f(z)$, whose corresponding Laurent series admit the same region of convergence, and thereby, must agree with each other by the uniqueness. So what can you tell from these together with the assumption that $f(z)$ is a bijection?

Answer 1

I think you meant $az$ for some $a \neq 0$ or $\frac b z$ for some $b \neq 0$.

Hints: By Picard's Theorem $0$ is not an essential singulairty of $f$. [In a neighborhood of an essential singularity the function attains 'most' of values infinitely many times, so it cannot be one-to-one]. If it is a pole than $|f(z)| \to \infty$ as $ z \to 0$ so $\frac 1{f(z)}$ becomes an entire function (after removing the singularity at $0$. It is also well known that the only one-to-one entire functions are of the type $ax+b$. You can finish the proof now.