The number of solutions of the equation z2+z¯=0 is

Let z = x + iy, so that z¯=x−iy.∴ z2+z¯=0⇒ x2−y2+x+i(2xy−y)=0Equating real and imaginary parts, we getx2−y2+x=0           (1)and 2xy−y=0⇒y=0 or x=12If y = 0, then (1) gives x2+x=0⇒ x=0 or x=−1If x = 1/2, then from Eq. (1),y2=14+12=34 or y=±32Hence, there are four solutions in all.