Solution:
SOLUTION
Let z = x + iy then z¯= x − iy
z2 = x2 − y2 + 2xyi
z2+z¯ = x2 − y2 + 2xyi + x − yi = 0
⇒(x2−y2+x)+(2xy−y)i=0
Step 2
Equating the real and imaginary terms on either sides
x2−y2+x=0 and 2xy−y=0
⇒y=0 or 2x=1 or x=1 / 2
case (i)
⇒if y=0 then x=0 or x=1
i.e.,z=0 or z=1
case (ii)
If x=1 / 2, then
$$ y^2=\large\frac{1}{4}+\frac{1}{2}=\frac{3}{4}$$
$$\Rightarrow\:y=\pm\large\frac{\sqrt 3}{2}$$
$$i.e.,z=\large\frac{1}{2}\pm\frac{\sqrt 3}{2}i$$
∴ There are four values for z