$f$ is bi-holomorphic between two disks $B(z_1,r_1), B(z_2,r_2)$ , continuous on $\partial B(z_1,r_1)$ and maps $f(z_1)=z_2$ then $f$ is linear

Question

$f$ is bi-holomorphic (one-to-one, onto, holomorphic in $B(z_1,r_1)$ , and $f^{-1}$ is also holomorphic) between two disks $B(z_1,r_1), B(z_2,r_2)$ , continuous on $\partial B(z_1,r_1)$ and map between $B(z_1,r_1)$ and $B(z_2,r_2)$ centers, i.e $f(z_1)=z_2$. I wish to show that $f$ is a linear map. This question appeared under the maximum principle subject, so I thought maybe it is possible to show that $f'(z)$ is constant map, which promises (using Taylor series) that $f=a_0+a_1 z$. However I didn't succeed to do so. I would also appreciate other ways to prove it, and hints for what can be deduced from the fact that $f$ maps between the disks' centers.

Answer 1

Hints: let $g(z)=z_2+\frac {r_2} {r_1} (z-z_1)$. Then $f\circ g^{-1}$ is a bi-holomorphic map on the second disk with fixes the center of the disk. By applying another linear tranformation we can reduce the question to the following: let $h$ be a bi-holomorphic map of the open unit disk which fixes the origin. Then $h(z)=e^{ia}z$ for some real number $z$. It is well known that all bi-hlomorphic maps of the open unit disk are of the form $z \to e^{ia} \frac {z-c} {1-\overline {c}z}$ for some real number $a$ and some $c$ in the open unit disk. The result follows from this.

Answer 2

You can assume both discs are the unit discs $B(0,1)$. Then the biholomorphic maps from $B(0,1)$ to $B(0,1)$ with $f(0)=0$ are the maps $z\mapsto cz$ with $|c|=1$ (so linear). This is a corollary of Schwarz's lemma.

You don't need the condition on the boundary (it is automatically satisfied).