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.


Linear fractional transformations send lines and circles to lines and circles. Our region is bounded by parts of a line and a circle; after transformation, it will still be bounded by parts of two lines and/or circles. The half-plane? Only one line.

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.

The closest we can come? We can map the half-disk to a quadrant with a linear fractional transformation.

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.

  • $\begingroup$ 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. $\endgroup$ – AstlyDichrar Jan 27 at 0:50

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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