Checking whether points form a polygon in complex plane

Question

If z^8=(z-1)^8 then the roots are 1) concyclic 2) form a polygonal 3)none

I found the roots to be 1+cot(k.pi/8) for k is a natural number and less than 8.

Then couldn't figure it out.

Answer 1

Consider the complex homography $$z \to f(z)=\frac{z}{z-1}$$

$f$ is an involution. So its inverse is equal to itself.

The solutions of the equation $Z^8=1$ are concyclic: they are on the unit circle. The solutions of your initial equation are located on the image of the unit circle by the complex homography $f$.

As the pole of $f$, namely $1$ belongs to the unit circle, the image of the the unit circle by $f$ is a line. Hence the solutions of the equation of your post lie on a line.