and $\lambda, \mu \in
{\mathbb{C}}^\ast$. The normal form is constructed similarly as above. In this case, when we obtain $x = y = 0$, the fact that $M \neq 0$ and $N = 0$ forces $r$ to be zero. In this case, we have only a single normal form, namely $$\beta = (0: \underbrace{1:i:0:i:-1:0:0:0:0}_M: \underbrace{0:0:0}_{x}:
\underbrace{0:0:0}_{y}: 0).$$

### Similarity Points. {#compactification:boundary:similarity}

The points $\beta = (0: M: x: y: r) \in B$ such that $M = 0$, $x \ne 0$ and $y
\ne 0$ are called *similarity points*. Since $x$ and $y$ are on the absolute conic, the complex dimension of the set of similarity points is $4$.

To compute normal forms of similarity points, we first apply rotations. As we saw in Subsection \[compactification:boundary:inversion\], right multiplication fixes $y$ and $r$ and can transform $x$ to $\gamma
{\thinspace}(1,i,0)^t$, and left multiplication fixes $x$ and $r$ and can transform $y$ to $\delta {\thinspace}(1, i, 0)^t$, with both $\gamma$ and $\delta$ in ${\mathbb{C}}^{*}$. Hence projectively we can always suppose that $\delta = 1$, so we can reduce any similarity point to one such that $x = \gamma {\thinspace}(1,i,0)^t$ and $y
=
(1,i,0)^t$. Then translations act transitively on $r$, thus we may get 