# Conformal mapping of a doubly connected domain onto an annulus

It is a well known theorem that any doubly connected domain can be conformally mapped onto an annulus.

Consider the simpler version :

Suppose $D$ is a bounded domain whose boundary is two non-intersecting circles. Then $D$ can be conformally mapped onto an annulus.

I believe that the proof of this should be easier, that the conformal map would be just a linear fractional transformation. However, I'm having trouble constructing it. Could someone help?

Thank you

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Side question: is a double connected domain doubly connected because there are two ways of getting from each point to another? – Mariano Suárez-Alvarez Sep 14 '10 at 18:56
@Mariano : Doubly connected usually means that the complement has two components (think of an annulus, for example), as opposed to simply connected. – Kalim Sep 14 '10 at 19:38

Suppose your outer circle is the unit circle $C_1=\{z:|z|=1\}$. All you have to do is to get a linear fractional transformation taking $C_1$ to itself and the inner circle $C_2$ to a circle centred at the origin. You might as well rotate $C_2$ so that its centre is on the real axis. Then $C_2$ meets the real axis in points $a$ and $b$ with $-1 < a < b < 1$. The required linear fractional transformation will have real coefficients and take $-1$, $a$, $b$, $1$ to $-1$, $-c$, $c$, $1$ where $0 < c < 1$. The typical linear fractional transformation fixing $-1$ and $1$ has the form $$f_u:z\mapsto\frac{z+u}{uz+1}.$$ We are considering $u$ real and for $u$ to take the unit disc to itself, $|u| < 1$. We need $f_u(a)=-f_u(b)$. This is a quadratic equation in $u$ with two real roots, one of which satsifies $|u| < 1$.