Complex Numbers #14

$\begin{aligned} z^2 & = \bar z & \small \color{#3D99F6} \text{Let } z = x+yi \\ (x+yi)^2 & = x-yi \\ x^2-y^2+2xyi & = x - yi \end{aligned}$

Equating the real parts and imaginarys part of both sides,

$\begin{cases} x(x-1) = y^2 & ...(1) \\ y(2x+1) = 0 & ...(2) \end{cases}$

From (2): $\begin{cases} y = 0 & \implies (1): \ x(x-1) = 0 \implies x = 0, \ 1 & \implies z = x +yi = \begin{cases} 0 \\ 1 \end{cases} \\ x = - \frac 12 & \implies (1): \ y^2 = - \frac 12\left(-\frac 12-1\right) \implies y = \pm \frac {\sqrt 3}2 & \implies z = x +yi = \begin{cases} - \frac 12 + \frac {\sqrt 3}2 \\ - \frac 12 - \frac {\sqrt 3}2 \end{cases} \end{cases}$

Therefore, there are $\boxed{4}$ values of $z$ satisfying the equation.