Nonidentity Möbius transformations are parabolic or conjugate to a dilation, with the projective trace invariant

Writing $M(z)=(az+b)\mathrm{/}(cz+d)$, the fixed-point equation is $c{z}^2+(d-a)z-b=0$ in the finite chart, together with the possibility that $\mathrm{\infty }$ is fixed. Therefore a nonidentity Möbius transformation has at most two fixed points.

If $M$ has exactly one fixed point $p$, then [L1] provides a Möbius map $T$ sending $p$ to $\mathrm{\infty }$. The conjugate $TM{T}^{-1}$ fixes $\mathrm{\infty }$, so it has the form $z\mapsto \alpha z+\beta$; uniqueness of the fixed point forces $\alpha =1$ and $\beta \mathrm{\ne }0$, so a further scaling conjugates it to $z\mapsto z+1$.

If $M$ has two fixed points $p\mathrm{\ne }q$, then [L1] provides a Möbius map sending them to $0$ and $\mathrm{\infty }$. The conjugate therefore fixes both $0$ and $\mathrm{\infty }$, hence has the form $z\mapsto \lambda z$ with $\lambda \in {\mathbb{C}}^{\times }\setminus \{1\}$.

Replacing a representing matrix $A$ by $tA$ multiplies both $\mathrm{tr}(A{)}^2$ and $\det A$ by $t^2$, so $\tau (M)=\mathrm{tr}(A{)}^2\mathrm{/}\det A$ is well defined on the projective class. For $z\mapsto \lambda z$ it equals $(\lambda +1{)}^2\mathrm{/}\lambda =\lambda +2+{\lambda }^{-1}$, and for $z\mapsto z+1$ it equals $4$.

Step 1.1 leaves only the one-fixed-point and two-fixed-point cases, and the preceding three steps identify those cases with the parabolic and dilation normal forms, the standard dilation-branch terminology, and the projective trace invariant.