# Conformal mapping of two circles onto an annulus

Given two circles: One with radius $1$ and center $i$, the other with radius $5$ and center $-i$, how can I find the Möbius transformation which maps them onto an annulus $1 < \lvert w\rvert< p$?

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Recall that Möbius transformations act 3-transitively on the points of the Riemann sphere. That is, we can find a Möbius transformation mapping any three points to any other three points. The algorithm to do this involves making three equations involving the four coefficients, and then solving.

In the present case, the desired Möbius transformation must map the points $(-i-5,0,2i)$ to the points $(-pi,-i,+i)$, so you can find the formula from there.

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A translation followed by a rotation will take the region to the "ring" between the circle centered at the origin of radius $1$ and the circle centered at $-2$ and radius $5$ (call it $D_1$).

The map $(z-1)/(z+1)$ will takes $D_1$ to the right half-plane with the circle centered in the real axis and diameter with extremes $1/2$ and $4/3$ removed (call it $D_2$).

On the other hand, the pap $f(z)=(z-1)/(z+1)$ takes the annulus $\{1<|z|<R\}$ to the right half-plane with the circle centered in the real axis and diameter with extremes $(R-1)/(R+1)$ and $(R+1)/(R-1)$ removed. Observe that the product of the extremes of the diameter is equal to $1$.

The dilation $\sqrt{3/2}\,z$ transforms $D_2$ to the right half-plane with the circle centered in the real axis and diameter with extremes $\sqrt{3/8}$ and $\sqrt{8/3}$ removed (call it $D_3$).

Finally, the map $g(z)=f^{-1}(z)=(1+z)/(1-z)$ transforms $D_3$ into the annulus $$\{1<|z|<\frac{\sqrt{8/3}+1}{\sqrt{8/3}-1}\}.$$

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Thanks. Very instructive. I had to draw some pictures to believe it! –  John Adamski May 17 '12 at 16:26