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?
Finding a conformal map from unit disk to half-plane
5
$\begingroup$
complex-analysis
conformal-geometry
2 Answers
1
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.
-
0Thanks! 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. – 2012-11-05
0
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....
-
0Thanks for the tip--I think your method gives the same function from the disk to the right half-plane as above. – 2012-11-05