The number of solution 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 get x2−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 (1),y2=14+12=34⇒y=±32 therefore solutions are 0,0,-1,0,12,±32Hence, there are four solution in all.