Determining a custom set in the complex plane Let $A(z) , B(z^2)$ and $C(z^3)$ 3 points on the complex plane.
Determine the set of points $A(z)$ so that $ABC$ is a rectangular triangle.
I tried the solving it by cases and found that $z=-1 +iy  (y>0)$, but I'm not sure if its correct so can I get some help
 A: The squares of the sides of $\triangle ABC$ are:
$$
\begin{align}
AB^2=|z^2 -z|^2 &= |z|^2 |z-1|^2 \\
BC^2=|z^3-z^2|^2 &=|z|^4 |z-1|^2 \\
CA^2=|z-z^3|^2 &=|z|^2|z-1|^2|z+1|^2
\end{align}$$
Dividing by the common factor $|z|^2|z-1|^2\,$, it follows that the squares of the  sides are proportional to $c^2=1\,$, $a^2=|z|^2\,$, and $b^2=|z+1|^2=(z+1)(\bar z+1)=|z|^2+2 \operatorname{Re}(z)+1\,$.
By Pythagoras' theorem the triangle is a right triangle iff one of those squares is equal to the sum of the other two, which leaves $3$ cases to consider:


*

*right angle at $A\,$ $\iff a^2=b^2+c^2$


$$\require{cancel}
\cancel{|z|^2}=\cancel{|z|^2}+2 \operatorname{Re}(z)+1 + 1 \;\;\iff\;\; \boxed{\operatorname{Re}(z) = -1}
$$


*

*right angle at $B\,$ $\iff b^2=c^2+a^2$


$$
\cancel{|z|^2}+2 \operatorname{Re}(z)+\bcancel{1} = \bcancel{1} + \cancel{|z|^2} \;\;\iff\;\; \boxed{\operatorname{Re}(z) = 0}
$$


*

*right angle at $C\,$ $\iff c^2=a^2+b^2$


$$
\cancel{1} = |z|^2 + |z|^2+2 \operatorname{Re}(z)+\cancel{1} \;\;\iff\;\; |z|^2 = -\operatorname{Re}(z) \;\;\iff\;\; \boxed{\left|z+\frac{1}{2}\right|^2=\frac{1}{4}}$$
Therefore the solution set is the union of two vertical lines and a circle in the complex plane.
