# Finding a conformal map from unit disk to half-plane

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?

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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.

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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

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

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

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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