Geometric images of complex numbers $z$ such that triangle with vertices $z, z^2,z^3$ is a right angled triangle.

If $(z,z^2,z^3)$ forms a nondegenerate right triangle then $z\ne0$, and $(1,z,z^2)$ forms a right triangle as well. (Note that $z$ is constant for a given triangle. The map $T_z: \> w\mapsto{1\over z} w$ is a similarity of the complex $w$-plane.) Conversely: If $(1,z,z^2)$ forms a nondegenerate right triangle then $z\ne0$, and $(z,z^2,z^3)$ forms a right triangle as well. We therefore look at nondegenerate triangles $(1,z,z^2)$. One has $$|z^2-1|^2=|z+1|^2|z-1|^2,\quad |z^2-z|^2=|z|^2\,|z-1|^2\ ,$$ hence all three sidelength squares have a factor $|z-1|^2\ne0$. We have to distinguish three cases:

(i) Right angle at $1$. By Pythagoras' theorem we then have $$|z-1|^2+|z^2-1|^2=|z^2-z|^2\ ,$$ or $$1+|z+1|^2=|z|^2\ .$$ This amounts to ${\rm Re}(z)=-1$, whereby the point $-1$ is excluded.

(ii) Right angle at $z$. By Pythagoras' theorem we then have $$|z-1|^2+|z^2-z|^2=|z^2-1|^2\ ,$$ or $$1+|z|^2=|z+1|^2\ .$$ This amounts to ${\rm Re}(z)=0$, whereby the point $0$ is excluded.

(iii) Right angle at $z^2$. By Pythagoras' theorem we then have $$|z^2-1|^2+|z^2-z|^2=|z-1|^2\ ,$$ or $$|z+1|^2+|z|^2=1\ .$$ This amounts to $\bigl|z+{1\over2}\bigr|^2={1\over4}$ (a circle), whereby the two points on the real axis have to be excluded.