# Möbius transformations

Teaching note: Möbius Transformations

This short note is written for those taking my second-year analysis course.

Möbius transformations are brilliant. They are conformal (=angle preserving) maps from the extended complex plane to itself.

Definition 1 (The extended complex plane). The extended complex plane is ${\overline{\mathbb{C}}:=\mathbb{C}\cup\{\infty\}}$, i.e. ${\mathbb{C}}$ together with an entirely new element, denoted ${\infty}$.

For now, we won’t extend the domains of ${+}$ and ${\cdot}$ to include ${\infty}$; so the extended complex plane has no more algebraic structure than the field of complex numbers.

Definition 2 (Möbius transformations). A Möbius transformation is a function ${f:\overline{\mathbb{C}}\longrightarrow\overline{\mathbb{C}}}$ such that

$\displaystyle f(z)=\left\{\begin{array}{ll}\frac{az+b}{cz+d}&z\neq-\frac{d}{c},\infty\\\infty&z=-\frac{d}{c}\\\frac{a}{c}&z=\infty,\end{array}\right.$

for some ${a,b,c,d\in\mathbb{C}}$ such that ${ad-bc\neq0}$.

The second and third cases in the definition above are important, but we specify a Möbius transformation simply by writing e.g.

$\displaystyle f(z):=\frac{az+b}{cz+d},$

and leave the other two cases implicit.

Nevertheless, such a Möbius transformation does not determine the quadruple ${(a,b,c,d)}$, but does determine the tuple ${[a;b;c;d]}$ in projective space. For a trivial example: the identity map ${z}$ is a Möbius transformation, equal to ${\frac{2z}{2}}$.

Example 1. Consider the Möbius transformation ${g:\overline{\mathbb{C}}\longrightarrow\overline{\mathbb{C}}}$ defined by ${g(z):=\frac{1}{z-1}.}$ Note that ${g}$ really is a Möbius transformation since ${-1\neq0}$. A few example evaluations: ${g(0)=-1}$, ${g(1)=\infty}$, and ${g(\infty)=0}$.

Proposition 3. The composition of two Möbius transformations is a Möbius transformation.

Proof: For ${i=1,2}$, consider a Möbius transformation ${f_{i}(z):=\frac{a_{i}z+b_{i}}{c_{i}z+d_{i}}}$, where ${a_{i}d_{i}-b_{i}c_{i}\neq0}$. The composition is $\displaystyle \begin{array}{rcl} f_{2}\circ f_{1}(z)&=&\frac{(a_{1}a_{2}+c_{1}b_{2})z+b_{1}a_{2}+d_{1}b_{2}}{(a_{1}c_{2}+c_{1}d_{2})z+b_{1}c_{2}+d_{1}d_{2}},\end{array}$ which is a Möbius transformation since $\displaystyle (a_{1}a_{2}+c_{1}b_{2})(b_{1}c_{2}+d_{1}d_{2}) - (a_{1}c_{2}+c_{1}d_{2})(b_{1}a_{2}+d_{1}b_{2}) = (a_{1}d_{1}-b_{1}c_{1})(a_{2}d_{2}-b_{2}c_{2})$ is not zero. $\Box$

Proposition 4. Möbius transformations are bijections.

Proof: Let ${f(z)=\frac{az+b}{cz+d}}$, with ${ad-bc\neq0}$. We find the candidate for its inverse, and check it really is. Define ${g:\overline{\mathbb{C}}\longrightarrow\overline{\mathbb{C}}}$
by

$\displaystyle g(w)=\frac{dw-b}{-cz+a}.$

We compose:

$\displaystyle \begin{array}{lll} g\circ f(z) &=& \frac{d\frac{az+b}{cz+d}-b}{-c\frac{az+b}{cz+d}+a}\\ &=& \frac{(ad-bc)z}{ad-bc}\\ &=& z, \end{array}$

the identity! The composition ${f\circ g}$ is the identity by the same calculation. $\Box$

Definition 5 (Möbius group). The set of Möbius transformations, equipped with composition, forms the Möbius group, denoted ${\mathbf{M}}$.

We are able to represent Möbius transformations as matrices, in the following way.

Definition 6. Define the map

$\displaystyle \begin{array}{rcl} \Phi:\mathbf{M} &\longrightarrow& \mathrm{PGL}_{2}(\mathbb{C})\\ \frac{az+b}{cz+d} &\longmapsto& \begin{pmatrix} a & b \\ c & d \end{pmatrix} \end{array}$

Naturally we wonder whether, under this representation of Möbius transformations as matrices, the composition of functions corresponds to the multiplication of matrices.

Proposition 7. ${\Phi}$ is an isomorphism.

Proof: It is clear that ${\Phi}$ is a bijection. For ${i=1,2}$, consider ${f_{i}\in\mathbf{M}}$ defined by ${f_{i}(z):=\frac{a_{i}z+b_{i}}{c_{i}z+d_{i}}}$, where ${a_{i}d_{i}-b_{i}c_{i}\neq0}$.
The composition is

$\displaystyle \begin{array}{rcl} f_{2}\circ f_{1}(z)&=&\frac{(a_{1}a_{2}+c_{1}b_{2})z+b_{1}a_{2}+d_{1}b_{2}}{(a_{1}c_{2}+c_{1}d_{2})z+b_{1}c_{2}+d_{1}d_{2}},\end{array}$

and we have

$\displaystyle \begin{array}{rcl} \Phi(f_{2})\Phi(f_{1}) &=& \begin{pmatrix}a_{2} & b_{2} \\ c_{2} & d_{2} \end{pmatrix}\begin{pmatrix} a_{1} & b_{1} \\ c_{1} & d_{1} \end{pmatrix}\\&=& \begin{pmatrix} a_{1}a_{2}+c_{1}b_{2} & b_{1}a_{2}+d_{1}b_{2} \\ a_{1}c_{2}+c_{1}d_{2} & b_{1}c_{2}+d_{1}d_{2} \end{pmatrix}\\&=& \Phi(f_{2}\circ f_{1}),\end{array}$

which shows that ${\Phi}$ is a homomorphism. $\Box$

Proposition 8. Möbius transformations act ${3}$-transitively on ${\overline{\mathbb{C}}}$.

Proof sketch: To see this, let ${w_{0},w_{1},w_{\infty}}$ be three distinct elements of the extended complex plane. We wish to choose ${[a;b;c;d]}$ such that

$\displaystyle f(z)=\frac{az+b}{cz+d}$

satisfies

$\displaystyle f(0)=w_{0},\,f(1)=w_{1},\,f(\infty)=w_{\infty}.$

Equivalently, such that the three equations

• ${\frac{b}{d}=w_{0}}$
• ${\frac{a+b}{c+d}=w_{1}}$
• ${\frac{a}{c}=w_{\infty}}$

hold. It suffices to satisfy $\displaystyle w_{\infty}c+w_{0}d=w_{1}c+w_{1}d,$ which can indeed be accomplished. $\Box$

1. Exercise: find the inverse of $f(z)=\frac{z-1}{z+1}$.