Set of points in complex plane

$\begin{aligned} \left(1+\frac 1z\right)^4 & = 1 = e^{2k\pi i} & \small \blue{\text{for }k=0,1,2,3} \end{aligned}$

$\begin{aligned} 1 + \frac 1z & = e^{\frac {k\pi} 2i} = \begin{cases} 1 & k = 0 \\ i & k = 1 \\ -1 & k = 2 \\ -i & k = 3 \end{cases} \end{aligned}$

$\dfrac 1z = \begin{cases} 0 \\ i - 1 \\ - 2 \\ -1 -i \end{cases} \implies z = \begin{cases} \infty & \small \red{\text{undefined}} \\ \dfrac 1{i-1} & = -\dfrac 12 - \dfrac i2 \\ \dfrac 12 & = -\dfrac 12 - 0i \\ - \dfrac 1{1+i} & = - \dfrac 12 + \dfrac i2 \end{cases}$

The three roots are colinear on the real line of $-\dfrac 12$ .

There are four fourth roots of $1$ . These are $1, -1, i, -i$ where $i = \sqrt{-1}$

Solving the equation using each of these roots we get :

$\newline$ $(A)\;\;1 +\Large\frac{1}{z}$ $= 1 \Rightarrow \text{No complex } z \text{ possible.}$

$\newline$ $(B)\;\;1 + \Large\frac{1}{z}$ $= -1\Rightarrow z = \Large\frac{1}{-2}$ $\Rightarrow \boxed{z = -\frac{1}{2}}$

$\newline$ $(C)\;\;1 + \Large\frac{1}{z}$ $= i\Rightarrow z = \Large\frac{1}{i - 1}$ $= \Large\frac{i + 1}{i^2 - 1}$ $= \Large\frac{i + 1}{-2}$ $\Rightarrow\boxed{z = \frac{-1}{2} - \frac{i}{2}}$

$\newline$ $(D)\;\;1 + \Large\frac{1}{z}$ $= -i\Rightarrow z = \Large\frac{1}{-i - 1}$ $= \Large\frac{i - 1}{-(i^2 - 1)}$ $= \Large\frac{i - 1}{2}$ $\Rightarrow\boxed{z = \frac{-1}{2} + \frac{i}{2}}$

We see that all the solutions have same real part. So they must lie on a line. Hence they are collinear.