0

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??

BAI
  • 2,565
dipali mali
  • 555
  • The map $f : z \mapsto (z+2)^2$ fulfills your condition but is not conformal, as $f'(0) = 0$. In general, given any domain $D$ I think it is easy to write a non-conformal map : $f : D \to \Bbb C \backslash (-\infty, 0]$. –  Jun 05 '17 at 08:06

2 Answers2

1

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.

sharding4
  • 4,917
kvicente
  • 369
0

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 $\log(1/e) = -1$.

eyeballfrog
  • 22,485