Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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 ?

share|improve this question
    
Try the Riemann Mapping Theorem. –  Neal Apr 22 '12 at 20:47
add comment

1 Answer 1

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).

share|improve this answer
    
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
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.