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.

I'm trying to find a conformal map $f$ from the open unit disk to the set $\mathbb{C}-[-1/4,-\infty)$ (I think this means the half-plane Re$(w)>-1/4$ with the properties $f(0)=0$ and $f'(0)>0$. I know that the mapping $$f(z)=\frac{i+z}{i-z}$$ returns the right half-plane Re$(w)>0$ from the open unit disk, but subtracting 1/4 from it doesn't satisfy $f(0)=0$. I can't seem to find a lot of other examples. Are there any other conformal maps that I should try?

share|improve this question
add comment

2 Answers

up vote 0 down vote accepted

Here's an outline; I'll leave the details to you:

The map you have will send the unit disc to a half plane. To get from a half plane to all of $\mathbb{C}$ minus a ray, postcompose with $z\mapsto z^2$. Now, to get the missing ray where you want it, rotate and translate.

Lastly, look at the pre-image of $0$. You can precompose with an automorphism of the disk sending $0$ to that point. Then all that's left is to check that, when you compose all these maps, the derivative is a positive number.

share|improve this answer
    
So the image of the right half-plane under $z^2$ is $\mathbb{C}$ without the ray from $0$ to $-\infty$, and translating this by -1/4 gives a conformal map that maps from the domain to the range that we want. So $f(z)=((i+z)/(i-z))^2-1/4$, and $f(-i/3)=0$. How would I rotate $z=-i/3$ on the unit disk to $0$? –  ff90 Nov 4 '12 at 4:45
    
I think that I would have to translate $-i/3$ to $0$, which after precomposing gives $z\mapsto (\frac{2i+3z}{4i+3z})^2-1/4$. But then if I'm not mistaken, $f'(0)$ is not $>0$.... –  ff90 Nov 4 '12 at 6:14
    
Sorry, that's not a rotation. I'll adjust my answer above. –  Brett Frankel Nov 4 '12 at 14:45
    
Thanks! Choosing a specific automorphism of the unit disk makes it work out and gives the simple-looking $f(z)=\frac{z}{(z-1)^2}$ as the answer. –  ff90 Nov 5 '12 at 0:38
add comment

Start with

$$L_1(z)=\frac{i}{2} \frac{z+1}{1-z}$$

Then rotate -90 degrees to get the right half plane. Then $z^2$ etc....

share|improve this answer
    
Thanks for the tip--I think your method gives the same function from the disk to the right half-plane as above. –  ff90 Nov 5 '12 at 0:40
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.