# Find a one-to-one conformal mapping from $D(0;1) \setminus \overline{D(1;1)}$ onto $D(0;1)$

I'm working on the conformal mappings of complex analysis. I can find conformal mappings from&onto simple domains but cannot find one from the abnormal domains.

Can any one give me a hint? Is there a rule that I can apply to find such mappings?

($D(z;r)$ means the open disk centered at $z$ with radius $r$.)

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I just realized your question has a little problem after noticing the answer given by @pritam. Actually $D(0;1)\setminus D(1;1)$ is not conformal to $D(0;1)$, and my answer gives a conformal map from $D(0;1)\setminus \overline{D(1;1)}$ to $D(0;1)$. – 23rd Dec 14 '12 at 17:18
I think the correct region should be the open region inside the unit disc and outside the circle centered at 1 with radius 1 i.e. $\{ |z|<1 \}$ and $\{|z-1|>1 \}$. – PAD Dec 14 '12 at 17:39

Hint: Note that the boundary of your region is a union of two circular arcs, which share two common end points. At a first step, you may choose a fractional linear transformation, which maps one end point to $0$ and the other end point to $\infty$. Then both circular arcs are mapped to rays with the same initial point $0$. Secondly, you may choose an appropriate power function to map the region bounded by the two rays to the upper half plane. Finally, you may use another fractional linear transformation to map the upper half plane to the unit disk.
I think there is no conformal map between your domains, because conformal mapping must preserve angle between two curves. If $P$ is one of the corner points of the domain $D(0,1)\setminus D(1,1)$, then the angle formed at $P$ is less than $\pi$ but the image of $P$ must be on the circle $C(0,1)$ and so the angle formed there is $\pi$; hence the angle is not preserved.