function with range ${\mathbb{C}\backslash (-\infty,o]}$

Question

let f be an analytic function defined on $D=\{z\in{\mathbb{C}:\vert{z}\vert \lt1}\} $ such that the range of $f$ is contained in the set $\mathbb{C}\setminus (-\infty,0]$

is $f$ is conformal mapping?

I know that analytic function is conformal at any point at which derivative is not zero.

Here if I will take example $f(z)=\log z$ on $D$ then it is conformal and satisfies given conditions but is it true in genral for all functions with given range??

Answer 1

If $U\subseteq \mathbb C$ and $U$ is open, then $f:U\rightarrow\mathbb C$ is conformal if and only if it is analytic everywhere in $U$ and $f'(z) \ne 0$ for all $z \in U$.

Your set $D\subseteq \mathbb C$ is open and $f$ maps from it to $\mathbb C$, so this should apply.

It's worth noting that $\log z$ does not satisfy your conditions. It is not analytic at all points in $D$ and its range is not in $\mathbb C \backslash (-\infty,0]$. This can be seen from the fact that $\log(1/e) = -1$.

Answer 2

As long as $f$ is holomorphic in the unit disc, and it's complex derivative is non-zero in the unit disc, then the unit disc will be conformally equivalent to the range of $f$ iff the range of $f$ is simply connected because of the riemann mapping theorem.