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

My try:
Let the complex number $z=x+yi$
Hence we need to find the locus of the point $(x,y)$ in the argand plane such that the points $(x,y)$, $(x^2-y^2, 2xy)$, $(x^3-3y^2x, 3x^2y-y^3)$
Now here I could apply the formula that the product of the gradient of two sides of triangle formed by these points is $-1$ but the method is a bit cumbersome.  Is there any other geometrical interpretation of it or any algebraic simplification further?
 A: 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.
A: An illustration to 
this answer.
For the most interesting part of the locus of $z$,
namely for the circle 
$\bigl|z+{1\over2}\bigr|^2={1\over4}$
with the two real points excluded,
this is an image of several right triangles:

and this is the locus of all triangles,
where $z$-points marked as red dots, 
$z^2$-points  marked as green dots
,$z^3$-points as blue dots
and the sides of the right triangles
painted in transparent orange:

A: I don't consider this to be a complete answer. However, I think that to share this idea is not a sin.
I've found at least one situation in the case of which we obtain a right triangle.

The imaginary axis is a locus of such complex numbers.

Poof
Consider the following figure where $z=re^{i\alpha}.$

The question is how to get a right triangle out of the points $z$, $z^2$, $z^3$, the red vectors. One solution is that we form a Thales triangle in a circle of radius $R$. In order to get there we need to solve the following two equations:
$$3\alpha-\alpha=\pi$$
and (notice that $R=\frac{r^3+r}2$)
$$(R-r)^2+r^4=R^2.$$
This latter equation is an identity and $\alpha=\frac{\pi}2$.

Example
Let $z=2e^{i\frac{\pi}2}$. In this case $R=5$ and we have the following configuration: $z=2e^{i\frac{\pi}2}, \ z^2=4e^{i\pi}, \ z^2=8e^{i\frac32\pi}$. Or

A: Let the squared-sides of the triangle be
$$\begin{align}
a^2 &\;:=\; |z^1-z^2|^2 \;=\; |z|^2 |1-z|^2 \\
b^2 &\;:=\; |z^2-z^3|^2 \;=\; |z|^4 |1-z|^2 \\
c^2 &\;:=\; |z^3-z^1|^2 \;=\; |z|^2 |1-z|^2 |1+z|^2
\end{align}$$
Trivially, we have a Pythagorean triple if $z=0$ or $z=1$. Otherwise, we can divide each of $a^2$, $b^2$, $c^2$ by $|z|^2 |1-z|^2$ to get, for $z = r(\cos\theta + i \sin\theta)$ with $r\neq0$ and $\theta\neq 0$,
$$a_\star^2 = 1 \qquad b_\star^2 = |z|^2 = r^2 \qquad c_\star^2 = |1+z|^2 =1+r^2+2r\cos\theta$$
so that $(a, b, c)$ is a (non-trivial) Pythagorean triple if and only if $(a_\star, b_\star, c_\star)$ is. Now, let's consider cases:

$$c_\star^2 = a_\star^2 + b_\star^2 \implies 1+r^2+2r\cos\theta = 1 + r^2 \implies \cos\theta = 0 \implies z = \pm p i \quad(p>0)$$

The diagram shows $z_1 = pi$ and $z_2 = -q i$. One may recognize that this is effectively the geometric mean configuration, which makes sense, since $|z_i^2|^2 = |z_i||z_i^3|$.


$$\begin{align}
a_\star^2 &= b_\star^2 + c_\star^2 \implies 1=r^2+1+r^2+2r\cos\theta \implies 2r\left(r+\phantom{r}\cos\theta\right) = 0 \implies r = -\cos\theta \\
b_\star^2 &= a_\star^2 + c_\star^2 \implies r^2 = 1 + 1 + r^2+2 r \cos\theta \implies 2\;\left(1+r\cos\theta\right) = 0 \implies r = -\sec\theta
\end{align}$$

Since $r$ must be non-negative, we restrict $z$ to Quadrants II and III. The diagram below shows typical solutions $z_3$ and $z_4$.

