= Solution
Subtracting the two defining equations gives
$$
y^2+z^2=(y-iz)(y+iz)=0.
$$
Consequently the variety is
$$
V(y-iz,x^2-z^2-1)
\ \cup\
V(y+iz,x^2-z^2-1).
$$
Each member is irreducible. Indeed, after imposing its linear equation, its <coordinate ring> is
$$
\mathbb C[x,z]/(x^2-z^2-1).
$$
The invertible linear substitution
$$
u=x-z,\qquad v=x+z
$$
turns this into
$$
\mathbb C[u,v]/(uv-1)\cong\mathbb C[u,u^{-1}],
$$
which is an <integral domain>. The two curves are distinct and neither contains the other, so they are precisely the <irreducible components of two intersecting complex hyperbolas>.
Back to article page
