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How can you tell if a conformal mappimg between regions is unique? I have a conformal mapping from {z : |z|<2, |Arg(z)|< pi/6} to {z : Re(z)>0, Im(z)<0} as

f(z) = -(iz^3 - 8)/(iz^3 + 8) but have no idea how to find out if the conformal mappimg is unique? Any help is much appreciated.

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  • $\begingroup$ for uniqueness (of conformal maps between simply connected domains) you need to fix the following pieces of data: $z_0$ in the original domain, $w_0$ in the target domain, $\theta$ an angle; then there is unique conformal $f$ from one domain onto the other with $f(z_0)=w_0, \arg f'(z_0)= \theta$; note that $|f'(z_0)|$ depends on the geometric shapes of the domains and the choice of the image point $w_0$ $\endgroup$ – Conrad Mar 25 at 22:18
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Such a conformal mapping will not be unique in general.

For example, suppose $A$ and $B$ are two nonempty, proper, simply connected open subsets of $\Bbb C$, and suppose that $f$ is a conformal mapping from $A$ to $B$.

By the Riemann mapping theorem, both $A$ and $B$ are conformally equivalent to the open unit disk $D$. In particular, there is a conformal mapping $g\colon A\to D$.

The open unit disk also has a whole bunch of conformal automorphisms, given by Möbius transformations, let $m$ be one such that is not the identity map. (A simple example, if we want to be concrete, is $m(z)=-z$.)

It follows that $f$ and $f\circ g^{-1}\circ m\circ g$ are distinct conformal mappings from $A$ to $B$.

(Indeed, for such domains, the set of conformal mappings from $A$ to $B$ actually forms an infinite group that is isomorphic to the group of conformal automorphisms of $D$.)

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    $\begingroup$ No need for a weird domain, for instance, for the complement to a generic finite subset of ${\mathbb C}$ consisting of at least 4 points, the group of conformal automorphisms is trivial. $\endgroup$ – Moishe Kohan Mar 25 at 22:26

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