Let $S$ be the set of all points $z$ in the complex plane such that $ \left(1+\frac{1}{z}\right)^{4}=1 $

Question

An entrance exam question

Let $S$ be the set of all points $z$ in the complex plane such that $$\left(1+\frac{1}{z}\right)^{4}=1$$ Then, the points of $S$ are
(A) vertices of a rectangle
(B) vertices of a right-angled triangle
(C) vertices of an equilateral triangle
(D) collinear

I had tried using $z = x+iy$ and simplifying the 4th-order equation, but it proved fruitless.

Answer 1

Hints: Taking of the $\space n$-th root has $\space n \space$ different solutions $\space \omega_k. \space$ The geometric interpretations of the points $\space \omega_k \space$ are the vertices of a regular $\space n$-gon whose center is at the origin.

Answer 2

These four points are on the image of the unit circle under a Moebius transformation. In this case the circle transforms to a line. One of the points becomes the point at infinity. The other three are collinear.