$x = \lambda w$ and $y = \mu v$ with $\lambda, \mu \in {\mathbb{C}}$. Hence the coordinates of an inversion point can be written as $(0:v w^t:\lambda w: \mu
v:r)$. Since $v$ and $w$ satisfy the quadratic equation $\langle v,v \rangle = \langle w,w
\rangle = 0$ (called the equation of the *absolute conic* in ${\mathbb{P}}^2_{{\mathbb{C}}}$), the complex dimension of the set of inversion points is 5 (one for $v$, one for $w$, one for $\lambda$, one for $\mu$, one for $r$). One can show that these are smooth points of the boundary, but we do not need this fact.

In order to compute normal forms, we first apply rotations. Multiplication from the right by a rotation of matrix $M'$ gives (see Remark \[remark:action\]) $$(0: v w^t M': \lambda {\thinspace}M' w: \mu {\thinspace}v: r),$$ so it leaves $v$ fixed. Being $M'$ an orthogonal matrix, it is in particular unitary, so it preserves both the scalar product and the Hermitian norm of $w$, and the action is transitive on vectors with $\langle w, w \rangle = 0$ and of the same Hermitian norm. Hence $w$ can be taken to a vector of the form $\delta {\thinspace}(1, i, 0)^t$, where $\delta \in
