Domain and range of Mobius transformation

Question

I read that the Mobius transformation $(z-i)/(z+i)$ is a biholomorphism from $\Re{z}>0$ to $B_1(0)$. How do I see this, and more generally, how do I determine the domains/ranges of general Mobius transformations?

Answer 1

Möbius transformations can always be considered to have domain and range the Riemann sphere $\mathbb{CP}^1 \cong S^2$. Möbius transformations are all maps from $\mathbb{CP}^1$ to themselves that take circles to circles (the restrictions of circles in $\mathbb{CP}^1$ to $\mathbb{C}$ are circles and lines, which we'll call "generalized circles." Actually, Möbius transformations are all orientation-preserving homeomorphisms of $\mathbb{CP}^1$ with this property).

This makes it quite straightforward to figure out where a Möbius transformation maps a half-plane or a disk (as is the question in your example here): it will always be another half-plane or disk. To determine which, it suffices to check where three boundary points of the first generalized circle go, as generalized circles are determined by three points. To determine which side of the generalized circle in the image your disk or half-plane went to, the fact that Möbius transformations preserve orientation determines it. A counter-clockwise triple of points in the first generalized circle as seen from the inside of the domain (a positively oriented triple) will be mapped to another positively oriented triple in the range, which determines the side uniquely.

Answer 2

You can use the following decomposition :

$$f(z)=\frac {az+b}{cz+d}=\frac {a}{c}+e \frac{1}{z+\frac {d}{c}}$$

(with $e=\frac{1}{c^2}(ad-bc)$). See paragraph "Composition of simple transformations" in this Wikipedia article.

which can be symbolized as a succession of 4 actions :

$$f=f_{4}\circ f_{3}\circ f_{2}\circ f_{1}.$$

where

• $u=f_1(z)$ is a translation (by $d/c$).

• $v=f_2(u)$ is the composition of an inversion and a symmetry with respect to the $x$ axis.

• $w=f_3(v)$ is a similitude (composition of a rotation and a homothety).

• $y=f_4(w)$ is (again) a translation (by $a/c$).

In this way, one can follow successively the transformations undergone by a figure because each one is understandable.

Here is an example with $f(z)=\frac{(1+i)z+(3-2i)}{iz+1}$ :

The initial figure is a caricatural fish (in black, around the origin, denoted P).

Then we have the successive transformations :

• $P \to P_1$ (red fish) by a $(0,-1)$ translation because $\frac{d}{c}=-i$;

• $P_1 \to P_2$ (magenta fish ; thick line) by the (non simple) transformation $z \mapsto 1/z$ ; see below for more explanations.

• $P_2 \to P_3$ (blue fish) by the combination of a rotation of $\approx -116°$ and homothety with ratio $\sqrt{5}$ because $e=-1-2i$.

• $P_3 \to P_4$ (cyan fish) by a translation by vector $(+1,-1)$ because $\frac{a}{c}=1-i$.

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Explanation about transformation $z \to 1/z$ :

This transformation can be in fact decomposed into two transformations : $z \to 1/\overline{z}$ which is the inversion (do you know this transformation ?) with the unit circle as its invariant circle followed by a conjugation of the result, which is the symmetry with respect to the $x$ axis. Let us have a look at the figure : the red fish is approximately "mirrored" by the inversion into the unit disk, giving the thin magenta which, in a second step, is "mirrored" into the $x$ axis in order to give the thick magenta fish.

Matlab program having generated the figure :

    clear all;close all;hold on;grid on;axis equal off;
    set(gcf,'color','w');
    LS='linesmoothing';LW='linewidth';
    a=1+i;c=i;d=1;b=3-2*i;e=(b*c-a*d)/c^2;
    f1=@(z)(z+d/c);
    f2=@(z)(1./z);
    f3=@(z)(e*z);
    f4=@(z)(z+a/c);
    f=@(z)((a*z+b)./(c*z+d));
    t=0:pi/200:2*pi;P=exp(i*t).*(cos(t).*cos(2*t));
    plot(P,'c',LS,'on');
    plot(exp(i*t),'m',LS,'on',LW,0.2);
    P=[P,P(1)];plot(P,'k',LS,'on')
    plot(f(P),'y','linewidth',4)
    plot([-2,4],[0,0],'k');
    plot([0,0],[-3,2],'k');
    P1=f1(P);plot(P1,'color','r',LS,'on',LW,2);
    P2=f2(P1);plot(P2,'color','m',LS,'on',LW,2);
    plot(conj(P2),'color','m',LS,'on',LW,0.2);% image by inversion
    P3=f3(P2);plot(P3,'color','b',LS,'on',LW,2);
    P4=f4(P3);plot(P4,'color','c',LS,'on',LW,2);
    n=50;
    T=[P(n),P1(n),P2(n),P3(n),P4(n)]+0.4i;
    text(real(T),imag(T),{'P','P_1','P_2','P_3','P_4'});