# Möbius transformation mapping

I would like to understand how to choose the right Mobius transformation. For example, the Mobius transformation that maps upper half plane onto the unit disk is:

$$z \rightarrow \frac{z-i}{z+i}$$

or

the Mobius transformation that maps the unit disk one to one onto the right half-plane is:

$$z\rightarrow\frac{1+z}{1-z}$$

I would like to understand how one gets to this Möbius transformations. Thank you!

• If you envision the complex plane as the Bloch sphere, then understand that Möbius transformations map "circles to circles", where these circles are on the Bloch sphere, then many times all you have to do is figure out where $0,1,$ and $\infty$ map to, and you're done. To be clear, the term "circles" here includes regular circles in the complex plane, as well as straight lines in the complex plane. Dec 13, 2018 at 15:09
• Would you be able to illustrate it for me with one of the examples above? I have never used the Bloch sphere before. Thank you! Dec 13, 2018 at 17:49

For your first example, because the boundary of the upper half-plane is a "circle" (in the Riemann sphere sense (sorry, Riemann sphere, not Bloch sphere)), and the boundary of the unit disk is a circle (plainly, but also in the Riemann sphere sense), we try to map the boundary of the one to the boundary of the other. We start with a generic Möbius transformation: $$w=\frac{az+b}{cz+d}.$$ Here $$a,b,c,d\in\mathbb{C}$$ and $$ad-bc\not=0.$$ Let's remember that $$z$$ is a point in the upper half-plane, and $$w$$ is in the unit circle. So if we consider the following table: $$\begin{array}{c|c}z &w \\ \hline 0 &b/d \\ \hline 1 &(a+b)/(c+d) \\ \hline \infty &a/c \end{array},$$ we need all three points in the second column to be unique points on the unit circle. How did I get the points for $$z?$$ The answer is this: they're the most straight-forward points on the real axis (the boundary of the upper half plane) to plug into the transformation. We can see right away that we need $$|b|=|d|,\; |a|=|c|,$$ and $$|a+b|=|c+d|,$$ in order for the $$w$$ points to be on the unit circle. There are a number of ways to do this, so let's try $$a=b=c=d=1:$$ $$\begin{array}{c|c|c}z &w &\text{actual} \; w \\ \hline 0 &b/d &1 \\ \hline 1 &(a+b)/(c+d) &1 \\ \hline \infty &a/c &1 \end{array}.$$ Clearly, this won't work: the $$w$$ have to be unique! Could we work backwards? How if we could map $$0\to 1, \; 1\to i, \; \infty\to-1?$$ Are there $$a,b,c,d$$ to make that work? Let's see: $$\begin{array}{c|c|c}z &w &\text{actual} \; w \\ \hline 0 &b/d &1 \\ \hline 1 &(a+b)/(c+d) &i \\ \hline \infty &a/c &-1 \end{array}.$$ So here $$b=d$$ and $$a=-c$$. To get $$i$$ out of $$\dfrac{a+b}{c+d}=\dfrac{a+b}{-a+b}$$ would require $$a+b=i(-a+b),$$ so (remembering that $$a,b\in\mathbb{C}$$) we would have \begin{align*} a(1+i)&=b(i-1) \\ a&=\frac{b(i-1)}{i+1}. \end{align*} We have one degree of freedom here, now. Let's just let $$b=1,$$ and see what falls out. We get $$a=\frac{i-1}{i+1}\cdot\frac{i-1}{i-1}=\frac{(i-1)(i-1)}{2}=\frac{-1-2i+1}{2}=-i.$$ Then, from before, we have $$d=1$$ and $$c=i,$$ so the mapping is $$w=\frac{-iz+1}{iz+1}.$$ We can see that $$ad-bc\not=0.$$ What about the interior? Let's see if $$z=i$$ maps to the interior of the unit circle. $$z=i$$ goes to $$(1+1)/(0)$$. Oops! Can't do that. We can see that $$z=-i$$ maps to $$w=0$$. So what's going on? We actually mapped the lower half-plane to the unit circle! The upper half-plane got mapped to everything outside the unit circle. What went wrong? Well, we had some freedom in choosing what to map $$0,1,\infty$$ to, so we must have done that wrong. Let's try going around the boundary of the unit circle the other way ($$0,1,\infty$$ got mapped to a counter-clockwise orientation on the unit circle before), and see if we can fix it: $$\begin{array}{c|c|c}z &w &\text{actual} \; w \\ \hline 0 &b/d &-1 \\ \hline 1 &(a+b)/(c+d) &? \\ \hline \infty &a/c &1 \end{array}.$$ We'll leave the middle one blank for now, but to get the clockwise orientation, it'll need to have a positive imaginary part. Let's try $$a=c=1$$. Now $$b=-d,$$ and we're going to need nonzero imaginary parts to get that third point right. The easiest thing to try is $$b=i, \; d=-i$$. Let's check the middle row: $$\frac{a+b}{c+d}=\frac{1+i}{1-i}\cdot\frac{1+i}{1+i}=\frac{1+2i-1}{2}=i,$$ which makes our transformation $$w=\frac{z+i}{z-i}.$$ Does this work? We check the interior again: $$z=i$$ should map to the interior of the unit circle. We get $$w=(2i)/(0),$$ which doesn't work again! So we try it the other way, with $$b=-i$$ and $$d=i$$, and you'll find that it works.
A simple answer is, when we want a map between unit disk and a right /left /upper /lower half-plane is considering the map $$w=\dfrac{1+z}{1-z}$$ as a base, which maps unit disk to right halfplane $$\{{\bf Re}\ w>0\}$$. With rotation $$\dfrac{\pi}{2}$$ of the range, we find a map $$i\dfrac{1+z}{1-z}:{\mathbb D}\to\{{\bf Im}\ w>0\}$$, other rotation makes $$i^2\dfrac{1+z}{1-z}:{\mathbb D}\to\{{\bf Re}\ w<0\}$$ and with another rotation we have the mapping $$i^3\dfrac{1+z}{1-z}:{\mathbb D}\to\{{\bf Im}\ w<0\}$$. Inverse of each of above maps, enable us to have a map from a half-plane to unit disk.
• $w=\frac{z+i}{z-i}$ maps upper half-plane to exterior of the unit disk. Let $\frac12+\frac{i}{2}\to2i-1$. Dec 14, 2018 at 2:59
• both are correct, $\frac{z-1}{1+z}=\color{red}{-}\frac{1-z}{1+z}$, each map flip the circle horizontally! Dec 14, 2018 at 7:40