# Mapping using mobius transformations

I had a fundamental question regarding mobius transforms. Suppose I want to map the unit circle to the upper half plane ($Im$ $z \geq 0$) on the complex plane.

I know mapping any three points on the line to three points on the circle is sufficient to uniquely determine a mobius transform.

My question is, are there an infinite number of such transforms that satisfy this mapping? i.e., if I choose $i$, $1$, and $-i$ on the circle to map to $0$, $1$, and $\infty$ on the line, I will get a particular mobius transform.

Now if I choose $i$, $\frac{1+i}{\sqrt2}$, and $1$ on the circle to map to $0$, $1$, and $\infty$ on the line, I will get another mobius transform.

Ultimately will both of these transforms be equivalent in mapping the unit circle to the half plane, even though they are different?

Also it seems I have only mapped the circle to the line, although I want to map the region within the circle to the upper half plane. How can this even work to map regions, although this is a standard technique?

Yes, there are infinitely many such transformations. You can always choose $3$ different points on the unit disk. As for your last question, it works because Mobius transformations map boundaries to boundaries. By mapping the unit circle to the real axis you know that the unit disk will go either to the upper half plane or to the lower half plane. So just check where does your transformation map $0$. By the way, the upper half plane is $(Imz\geq0)$ and not $(Rez\geq0)$.
• Corrected. So if I check and the transform maps $0$ to the lower half plane, then how can I ensure I find a transform that maps it to the upper half plane? – RagingStormlight Sep 17 '18 at 10:15
• Then just multiply your transformation by $-1$. If $x+iy$ is in the lower half plane then $-(x+iy)$ is in the upper half plane. – Mark Sep 17 '18 at 10:49