Geometric interpretation of complex number. 
Let $z \in \mathbb{C}, z^2, z^3$ be the verticies of a right triangle. Find the geometric images of $z$.


I did not understand the question but I guess it want me to find the figure formed by $z$ under these constraints.
Let $z = x + iy$, then $z^2 = x^2 - y^2 +2xyi$ and $z^3  = x^3 - iy^3 + 3x^2iy - 3xy^2$ 
It forms a triangle, So, $|z - z^2|^2 + |z^3 - z^2|^2 = |z - z^3|^2$ by Pythagoras theorem.
But when I put the values in I get polynomial in two degree with 6th degree with many extra terms, in short a complete mess. 
How can I do this  question without actually explanding the Pythogoras theorem ?

Hint: let $r = |z| \in \mathbb{R}^+\,$, so that $\bar z = r^2 / z\,$. 

It forms a triangle, So, $|z - z^2|^2 + |z^3 - z^2|^2 = |z - z^3|^2$ by Pythagoras theorem.

This assumes that the hypotenuse is $z\,z_3\,$, and the following works on this assumption. The other cases would follow similarly.
Rewriting it in terms of conjugates:
$$
\require{cancel}
\begin{align}
0 & = |z - z^2|^2 + |z^3 - z^2|^2 - |z - z^3|^2 \\ & = (z - z^2)(\bar z - \bar z^2) + (z^3 - z^2)(\bar z^3-\bar z^2) - (z - z^3)(\bar z -\bar z^3) \\
 & = \bcancel{r^2} + r^4 - z \bar z^2 - \bar z z^2 + \cancel{r^6} + r^4 - z^3 \bar z^2 - \bar z^2 z^3 - \bcancel{r^2} - \cancel{r^6} + z \bar z^3 + \bar z z^3 \\
 & = r^2 (z^2 + \bar z^2) - (r^2+r^4) (z+\bar z) +2 r^4 \\
 & = r^2 \left(z^2+ \frac{r^4}{z^2}\right) -r^2(r^2+1)\left(z + \frac{r^2}{z}\right) + 2 r^4 \\
 & = r^2\left( \left( z + \frac{r^2}{z}\right)^2 - \bcancel{2 r^2} \right) -r^2(r^2+1)\left(z + \frac{r^2}{z}\right) + \bcancel{2 r^4} \\
 & = r^2 \left( z + \frac{r^2}{z}\right) \left( z + \frac{r^2}{z} - r^2 -1 \right) \\
 & = r^2 \left( z + \frac{r^2}{z}\right)\,\frac{(z-1)(z-r^2)}{z}
\end{align}
$$
The latter can be easily solved and, discarding the degenerate triangles given by $z=0,1\,$, the solutions are $z=\pm i \,r\,$ with $r \in \mathbb{R}^+\,$, which is the same as $z = \lambda i$ with $\lambda \in \mathbb{R} \setminus \{0\}\,$.
