# Möbius transformation in the complex plane.

Assume that $U$ be a line in the complex plane.

And assume a Möbius transformation $\phi$ sends $U$ again to a line.

How can I classify all such $\phi$?

I want to write my ideas. But, I cannot do anything. Please explain. I saw this question in a textbook as an exercise. This seems so interesting to me. I just want to learn this question.

Well, every Möbius transformation maps generalized circles to generalized circles. In particular, the line $U$ will be sent to either a line or a circle, and it will be sent to a line if and only if some point on the line is sent to the point at infinity. There are two possibilities, then, for what form such a transformation will have.

Option 1: If the point at infinity (which one can consider as a point on every line in the plane) is sent to the point at infinity, then we are dealing with a map of the form $\phi(z)=\zeta z+\xi,$ where $\zeta,\xi$ are complex constants with $\zeta\ne 0.$

Option 2: If some other point on the line $U$ is sent to the point at infinity (say the point $z_0$), then we are dealing with a map of the form $$\phi(z)=\frac{\alpha z+\beta}{z-z_0},$$ where $\alpha,\beta$ are complex constants such that $\alpha z_0+\beta\ne 0.$

Here is the outline of a proof that the above classification covers all such $\phi,$ using results from the link you posted:

To begin with, we know we are dealing with Möbius transformations, which are of the form $$\phi(z)=\frac{az+b}{cz+d},$$ where $a,b,c,d\in\Bbb C$ such that $ad-bc\ne0.$ By Theorem 3.4, we know that any such transformation will map our line $U$ into either a circle or a line. We can actually say more than Theorem 3.4 says. If we consider the point at infinity to be a point on every line, then we can (and should) show that Möbius transformations map circles and lines into and onto circles and lines. To prove this, we can use Lemma 3.2 (in particular, the formula for the inverse of a Möbius transformation), together with the definitions of arithmetic in $\hat{\Bbb C}=\Bbb C\cup\{\infty\}.$

If $\phi(z)=\zeta z+\xi$ for some $\zeta,\xi\in\Bbb C$ with $\zeta\ne0,$ then we can easily write $$\phi(z)=\frac{az+b}{cz+d}$$ with $a=\zeta,b=\xi,c=0,d=1,$ so that $ad-bc=\zeta\ne0$ (so we are dealing with a Möbius transformation), and since $c=0,$ then $\phi(\infty)=\infty$ by Lemma 3.8. Since there are points on $U$ arbitrarily far from the origin, then it can (and should) be shown that we can find $z\in U$ with $z\ne\infty$ such that $|\phi(z)|$ is as large as we like, meaning that $\phi(U)$ is not bounded, so cannot be a circle, and so must be a line. Hence, for all nonzero $\zeta\in\Bbb C$ and all $\xi\in\Bbb C,$ the transformation $\phi(z)=\zeta z+\xi$ is a Möbius transformation that will map the line $U$ to a line.

If $\phi(z)=\dfrac{\alpha z+\beta}{z-z_0}$ for some $z_0\in U,$ $z_0\ne\infty,$ and so $\alpha,\beta\in\Bbb C$ with $\alpha z_0+\beta\ne 0,$ then we have $$\phi(z)=\frac{az+b}{cz+d}$$ with $a=\alpha,b=\beta,c=1,d=-z_0,$ so $ad-bc=-\alpha z_0-\beta=-(\alpha z_0+\beta)\ne0$ (so we are dealing with a Möbius transformation), and since $c\ne 0,$ then $\phi(z_0)=\phi(-\frac dc)=\infty$ by Lemma 3.8. Since there are points on $U$ arbitrarily close to $z_0,$ then it can (and should) be shown that we can find $z\in U$ with $z\ne z_0$ such that $|\phi(z)|$ is as large as we like. By the same sort of reasoning as above, we see that for all $z_0\in U$ with $z_0\ne\infty$ and for all $\alpha,\beta\in\Bbb C$ such that $\alpha z_0+\beta\ne 0,$ the transformation $\phi=\dfrac{\alpha z+\beta}{z-z_0}$ is a Möbius transformation that will map the line $U$ to a line.

All that is left is to show that the only Möbius transformations that will map the line $U$ to a line are those which can be written in the forms described above. So, we start with a general Möbius transformation $$\phi(z)=\frac{az+b}{cz+d}$$ (where $a,b,c,d\in\Bbb C$ such that $ad-bc\ne 0$) and suppose that $\phi(U)$ is a line. Since $\infty\in\phi(U),$ then one of the following must be true:

• There is some $z_0\in U$ with $z_0\ne\infty$ such that $\phi(z_0)=\infty.$
• $\phi(\infty)=\infty.$

In the first case, note that we cannot have $c=0,$ for if so, then since $ad-bc\ne 0,$ we have $d\ne 0,$ so that $\phi(z_0)=\frac{az_0+b}d\in\Bbb C,$ contradicting our assumption that $\phi(z_0)=\infty.$ Since $c\ne 0,$ let $\alpha=\frac ac,\beta=\frac bc,$ so $\alpha,\beta\in\Bbb C,$ and dividing numerator and denominator by $c,$ we can rewrite $$\phi(z)=\cfrac{\alpha z+\beta}{z+\frac dc}.$$ Now, $$\infty=\phi(z_0)=\cfrac{\alpha z_0+\beta}{z_0+\frac dc},$$ so we must have $z_0+\frac dc=0,$ as otherwise $\phi(z_0)\in\Bbb C.$ Thus, $\frac dc=-z_0,$ so that $$\phi(z)=\frac{\alpha z+\beta}{z-z_0},$$ and $$\alpha z_0+\beta=-\frac{ad}{c^2}+\frac bc=-\frac{ad-bc}{c^2}\ne 0.$$

In the second case, we must have $c=0,$ as otherwise, we have by Lemma 3.8 that $\phi(\infty)=\frac ac\in\Bbb C.$ Hence, since $ad=ad-bc\ne 0,$ we can conclude that $a,d\ne 0,$ so letting $\zeta=\frac ad$ and $\xi=\frac bd,$ we have that $\zeta,\xi\in\Bbb C$ with $\zeta\ne 0,$ and we can show that $\phi(z)=\zeta z+\xi.$