Complex number such that affixes of itself, its squared and its cube form a right triangle. $z,z^2,z^3$ form a right triangle in argand plane $\DeclareMathOperator{\aff}{Aff}$
Here is a funny problem  I stumbled upon recently
I call the affix of a complex number the corresponding point in the argand plane, that is
$$\aff(\mathrm{i})=(0;1)$$
Find all $z \in \mathbf{C}$ such that the affixes of $z,z^2\ \& \ z^3$ form a right triangle.
I decided to Set $A=\aff(z),B=\aff(z^2),C=\aff(z^3)$
I have proven that :

*

*If $z$ is a solution, then $z\neq 0,\  z\neq 1, \ \& \ z\neq-1$

*If the triangle is right angled at A, $z$ must be a non zero imaginary number.

*If the triangle is rigth angled at B, $\aff(z)$ has to lie on the vertical line defined by $x=(-1)$, except for $z=(-1)$.

(I'll post my proof when I get some time to write it in $\mathrm{\LaTeX}$ soon.)
I'm down to the third case : I have proven that if the triangle is right angled at $C$, then :
$$\Re \left(\frac{1+z}{z}\right)=0$$
Thanks to some experimentation, I know that $\aff(z)$ should lie on the circle centered at $(-\frac12;0)$ with radius $\frac12$, with $z$ again respecting the first condition above.
Now,how do I prove this ?
Thanks for the help.
Here is an animation.

 A: Partial Solution:
If $z=\cos t+i\sin t,$
$$m_{AB}=\dfrac{\sin2t-\sin t}{\cos 2t-\cos t}=-\cot\dfrac{3t}2\text{ if }\sin\dfrac t2\ne0$$
Similarly,
$$ m_{BC}=-\cot\dfrac{5t}2, m_{CA}=-\cot2t$$
Now we need at exactly one of the products $m_{AB}\cdot m_{BC},m_{BC}\cdot m_{CA},m_{CA}\cdot m_{AB}$  to be $=-1$
A: HINT
Angle between vectors $\vec{KM}, \vec{LM},$ where the points have complex coordinates $z_K, z_L, z_M,$ is equal to
$$\arg{\frac{z_M-z_L}{z_M-z_K}}.$$
In the present exercise we want
$$\arg{\frac{z-z^2}{z-z^3}}=\frac{\pi}{2}+k\pi,\;k\in \mathbb{Z}$$
or
$$\arg{\frac{z^2-z}{z^2-z^3}}=\frac{\pi}{2}+k\pi,\;k\in \mathbb{Z}$$
or
$$\arg{\frac{z^3-z^2}{z^3-z}}=\frac{\pi}{2}+k\pi,\;k\in \mathbb{Z}$$
A: Oh dear,
It turns out I'm too tired to think, I found the following :
let's set $z=a+ib$ as in the question
then :
$$\frac{z+1}{z}=\frac{(a+1)+ib}{a+ib}=\frac{(a+1)+ib)(a-ib)}{a^2+b^2}$$
From which one gets
$$\frac{z+1}{z}=\frac{a+a^2+b^2-ib}{a^2+b^2}=\frac{a+a^2+b^2}{a^2+b^2}-\frac{ib}{a^2+b^2}$$
now, because this is an imaginary number we get
$$C(a,b)=a+a^2+b^2=0$$
which is the equation of an ellipse because it is a polynomial of two variables, of degree two. We see that there is no division, therefore, this is the equation of a circle
Let's rewrite it to see the circle's centre and radius :
$$C(a,b)=a+a^2+(b-0)^2$$
That is with two canonical forms
$$C(a,b)=\left(a+\frac{1}{2}\right)^2+(b-0)^2-\frac14=0$$
But then we get
$$\left(a-\left(-\frac{1}{2}\right)\right)^2+(b-0)^2=\frac14$$
We just need a square after the equal sign to get the radius
$$\left(a-\left(-\frac{1}{2}\right)\right)^2+(b-0)^2=\left(\frac12\right)^2$$
which is the equation of the circle mentioned in the question / OP.
Please note however that $(-1)$ and $0$ both belong to this circle.
