I'm looking through an example of finding a biholomorphic map $f$ that maps from the unit disc $\mathbb{D}$ to $D_R(c) = \{z \in \mathbb{C}: |z-c| < R\}$ such that $f(0) = c$, $f'(0) > R$. I am having trouble understanding the part where they show that this map is unique.

It says that we should assume $h$ is any map with the same properties as $f$. Then $h^{-1}\circ f$ is a map from $\mathbb{D}$ to $\mathbb{D}$, $h^{-1}\circ f(0) = 0$ and that $h^{-1}\circ f$ is a rotation. That part I understand. I think. But then it goes on to say that since $(h^{-1}\circ f)'(0) = f'(0) \frac{1}{h'(0)} > 0$, we must have $f = h$. I have no idea why $(h^{-1}\circ f)'(0) > 0$ means that $f = h$...

  • 1
    $\begingroup$ If $g$ is a rotation, what is its derivative? (The point is that a rotation with positive derivative is the identity.) $\endgroup$ Jan 3, 2017 at 13:15
  • $\begingroup$ Oh. I suspected it was something about the rotation that I wasn't taking into consideration. Thank you! $\endgroup$
    – frej.mh
    Jan 3, 2017 at 13:23


You must log in to answer this question.

Browse other questions tagged .