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 ?

If the vertex at the right angle is $z^2$, then $\dfrac{z^3-z^2}{z-z^2}$ must be purely imaginary. Thus $\dfrac{z^2-z}{1-z} = -z$ is purely imaginary. Thus $z = ri$ for some real $r$.

Let $z = re^{i\theta}$ be the vertex at the right angle. From the figure we have 
\begin{align*}
AB^2 &= r^2 + r^4 -2r^3 \cos \theta \\
BC^2 &= r^4 + r^6 - 2r^5 \cos \theta\\
AC^2 &= r^2+r^6 - 2r^4 \cos 2\theta
\end{align*}
Using $BC^2 = AB^2 + AC^2$, we gave
$$ (r^2 + r^4 -2r^3 \cos \theta)+ (r^2+r^6 - 2r^4 \cos 2\theta) = r^4 + r^6 - 2r^5 \cos \theta$$
Simplifying we get
$$1 - r\cos\theta -r^2\cos 2\theta + r^3 \cos \theta = 0$$
When $\theta = 90^\circ$, we have 
$1 - r\cos\theta -r^2\cos 2\theta + r^3 \cos \theta  = 1+r^2 > 0$ and when $\theta = 180^\circ$, we have $1 - r\cos\theta -r^2\cos 2\theta + r^3 \cos \theta  = (1+r)(1-r^2)$ When $r > 1$, this quantity is negative and hence there is a $\theta$ that $\dfrac{\pi}{2} < \theta < \pi $ for which the above equation is satisfied. Hence there is a right angled triangle with $z$ as the vertex at the right angle. An example is shown below.
Similar situation happens when the vertex at right angle is $z^3$. 

