ISI B.Math Entrance (4)

$\displaystyle \bar{z} = z^2$

Taking conjugate,

$\displaystyle \large \bar{ \bar{z} } =( \bar{z })^2$

$\displaystyle \large z = ( \bar{z})^2 = ( z^2)^2 = z^4$

$\displaystyle \large \implies z( z^3 - 1 ) = 0$

Therefore , $\displaystyle z = 0,1, \omega , ( \omega )^2 ~~$ are the solutions. ( where $\omega$ is complex cube root of unity )

Set $z=x+iy$ , to get the set of equations

$\displaystyle \begin{cases} 2xy=-y \\ x^2-y^2=x \end{cases}$

Solving gives $\boxed{4}$ solutions. $\left[\displaystyle (0,0),(1,0),(-\dfrac{1}{2},\dfrac{\sqrt{3}}{2}),(-\dfrac{1}{2},-\dfrac{\sqrt{3}}{2})\right]$