# Existence of fractional linear map from right half unit disk into upper half plane

Let $$R=\{z\in\mathbb{C}:|z|<1,\text{Re}(z)>0\}$$, which is the right half of the unit disk. Does there exist a fractional linear map, that is, $$T:R\mapsto\mathbb{H}$$ such that $$T(z)=\frac{az+b}{cz+d}$$?

I don't think such a map does exist because we need to use $$z^2$$ as an intermediate step, and so $$T$$ can't have the form above. But I don't know how to prove it, it might be false for all I know.

More generally, how do we know what regions can be mapped from one to another using fractional linear transformations? For example, we could use a fractional linear transformation to map the unit disk into the upper half plane. But why can't it be done for this region? It feels like it should be possible but I haven't got a sensible result.

I thought it would be a sensible start to set $$T(-i)=0$$, $$T(0)=-1$$ and $$T(1)=1$$. I found $$T(z)=\frac{z+i}{(1+2i)z-i}$$ from these three points, but it is not mapping the curve $$\gamma(t)=e^{it}$$ for $$t\in(-\frac{\pi}2,\frac{\pi}2)$$ into the positive real axis.

To look at it another way, complex analytic functions are conformal (angle-preserving) anywhere the derivative is nonzero. For a linear fractional transformation? That's everywhere, except for whatever point gets sent to $$\infty$$. In this case, our region's boundary has two right angles in it. We can send one of them off to $$\infty$$, but the other will still be a right angle somewhere in the plane. There's no way we can force it to a half-plane unless we apply something that's not conformal (because of a zero derivative) at that point.
Your "sensible start"? That actually maps the circle passing through $$0$$, $$1$$, and $$-i$$ to the real axis. Needless to say, that's not the same as the unit circle, except at the two points of intersection. If you wanted to map a portion of the unit circle to a portion of the real axis with a LFT, you would have to choose three points on the circle to map to three points on the line 0 at which point we would be mapping the whole circle to the whole line.
• Thank you for your answer, it's great and clarified things a lot for me. It flew over me that $0$, $1$ and $-i$ are concyclic, though it's painfully obvious now that you said it. – AstlyDichrar Jan 27 at 0:50