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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
up vote 2 down vote accepted

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).

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Thank you. What do you mean by 'doubly connected'? Couldn't I find a conformal map i the sense that it is a holomorphic map into the unit disc with derivative different from zero (also non bijective)? – balestrav Apr 22 '12 at 21:08
@balestrav, multiply connected regions and concentric slit domains are briefly treated in Ahlfohrs, pages 243-253. Page 249, exercise 1, is about the ratio of radii. In this setting, doubly connected just means homeomorphic to an open annulus. – Will Jagy Apr 22 '12 at 22:11
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

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