For all time-like vectors $z$ the set $C(z)$ is a convex cone, i.e. for all $x,y\in C(z)$ and all $t\in[0,1]$: $(1-t)x+ty\in C(z)$. $Z$ is not convex!
Suppose $x,y\in C(z)$, then $\la x,x\ra < 0$, $\la y,y\ra < 0$ and $\la x,y\ra < 0$ and thus
$$
\la(1-t)x+ty,(1-t)x+ty\ra
=(1-t)^2\la x,x\ra+2t(1-t)\la x,y\ra+t^2\la y,y\ra
< 0,
$$
i.e. $(1-t)x+ty\in Z$ and since $\la(1-t)x+ty,z\ra=(1-t)\la x,z\ra+t\la y,z\ra < 0$: $(1-t)x+ty\in C(z)$. Actually, by the wrong Cauchy-Schwarz inequality:
$$
\la(1-t)x+ty,(1-t)x+ty\ra < -((1-t)\Vert x\Vert+t\Vert y\Vert)^2~.
$$
2. Suppose $z$ is a unit vector and $u\perp z$ satisfies $\Vert u\Vert < 1$. Then $z+u$ and $-z+u$ are time-like and
$$
\tfrac12(z+u)+\tfrac12(-z+u)=u
$$
is space-like. Hence $Z$ is not convex.