Find all the $z$ such that $(1+\frac{1}{z})^{4}=1$

Question

Question

if $S$ be the set of solution of $$\bigg(1+\frac{1}{z}\bigg)^{4}=1$$ then prove that the points are co-linear.

Attempt $\bigg(1+\frac{1}{z}\bigg)^{4}=1$ $\implies z^4+4z^3+6z^2+4z+1=z^4$ $\implies 4z^3+6z^2+4z+1=0$ $\implies (2z+1)(2z^2+2z+1)=0$

Answer 1

Just take the fourth root both sides to get $$1+\frac1{z}=\pm1,\pm i$$ $$\frac1{z}=-2,0,-1\pm i$$ $$z=-\frac12,-\frac12\pm \frac12i$$ Which are colinear solutions.

Answer 2

Hint Simpler soultion:

Let $w=1+\frac{1}{z}$. Then $$w^4=1 \Rightarrow w=1,-1,i,-i$$

$w=1$ is not possible, the other three lead to $$\frac{1}{z}=w-1 \Rightarrow z=\frac{1}{w-1}$$

Answer 3

$$ \Big((1+{1\over z})-1\Big)\Big((1+{1\over z})+1\Big)\Big((1+{1\over z})+i\Big)\Big((1+{1\over z})-i\Big)=0$$

So $z_1 =-{1\over 2}$ and $\displaystyle z_{2,3} ={-1\pm i\over 2}$

Since $$Re(z_i) = -{1\over 2}$$ for all $i$ the solution are collinear, all the solution are on a line paralle to imaginary axsis.