# Conformal mapping into the unit disc

Is there a bijective conformal mapping from $A=\mathbb{C}-[1,\infty[$ into the unit open disc?

I thought that I could translate A into $\mathbb{C}-[0,\infty[$ then consider $f(z)=z^{1/2}$ which should be defined since i cut a semiline. Now I should obtain the upper half-space and mao this into the unit disc. Is this correct? What about the same question from $B=\mathbb{C}-[0,1]$? In this case it coudn't be bijective since B is not simply connected right? But what could be a conformal map ?

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Try the Riemann Mapping Theorem. –  Neal Apr 22 '12 at 20:47

For the first question, your idea is correct. As for the second, since $D=\mathbb{C}\setminus[0,1]$ is doubly connected, it is conformaly equivalent to an annulus. To find a conformal mapping from $D$ to an annulus, consider the function $z+1/z$ on $\{0<|z|<1\}$ (or its inverse).
By doubly connected I mean that the complement in the Riemann sphere has two connected components: $\{\infty\}$ and $[0,1]$. –  Julián Aguirre Apr 23 '12 at 8:34