Question

The number of complex numbers which are conjugate of their own cube​

Answer 1

The maximum number of complex number is "2".

Step-by-step explanation:

Let Z = x + iy be the complex number of Z.

Where, x is called real part of Z and y is called imaginary part of Z.

The conjugate of complex number = x - iy

$(x -iy)^{3}$ = $x^{3}$ - $(iy)^{3}$ -3x(iy) {x - iy}

=  $x^{3}$ + i·$y^{3}$  - i(3xy) - 3·$x^{2}$·y

[ ∵ $i^{2}$ = - 1]

Clearly, the maximum number of complex number is "2".