Solve the equation, `z^(2) = bar(z)`, where z is a complex number.

1.	Solve the equation, `z^(2) = bar(z)`, where z is a complex number.
Answer» Let z = x + iy. Then, `z^(2) = bar(z) rArr (x+iy)^(2)=x - iy` `rArr" "(x^(2)-y^(2))+2ixy = x - iy." "...(i)` Equating real parts and imaginary parts on both sides of (i) separately, we get `x^(2)-y^(2)=x" "...(ii) and 2xy = -y" "...(iii)`. From (iii), we get `2xy + y = 0 rArr y(2x + 1) = 0 rArr y = 0 or x = (-1)/(2)`. Case I When y = 0. Putting y = 0 in (ii), we get `x^(2)-x = 0 rArr x(x-1) = 0 rArr x = 0 or x = 1`. `therefore" "(x = 0, y = 0)or(x = 1, y = 0)` Thus, `z = (0+i0) or z = (1 + i0)`. Case II When `x = (-1)/(2)`. Putting `x = (-1)/(2)` in (ii), we get `((-1)/(2))^(2)-y^(2)=((-1)/(2)) rArr y^(2)=((1)/(4)+(1)/(2))=(3)/(4) rArr y = +-(sqrt(3))/(2)`. `therefore" "(x=(-1)/(2),y=(sqrt(3))/(2))or(x=(-1)/(2),y=(-sqrt(3))/(2))`. Thus, `z=((-1)/(2)+(sqrt(3))/(2)i)or z=((-1)/(2)-(sqrt(3))/(2)i)`. Hence, `z=0,1((-1)/(2)+(sqrt(3))/(2)i)and((-1)/(2)-(sqrt(3))/(2)i)` are the required roots of the given equation.

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