Browse Questions

# Solve the equation $\;z^{2} = \overline{z}\;$ where $\;z=x+iy$

$(a)\;\large\frac{7}{2} \pm \large\frac{\sqrt{3}}{2}\qquad(b)\;-\large\frac{1}{2} \pm \large\frac{\sqrt{3}}{2}\qquad(c)\;-\large\frac{3}{2} \pm \large\frac{\sqrt{5}}{2}\qquad(d)\;-\large\frac{5}{2} \pm \large\frac{\sqrt{3}}{2}$

Answer : $\;-\large\frac{1}{2} \pm \large\frac{\sqrt{3}}{2}$
Explanation :
$\;z^{2} = \overline{z}\;$
$(x+iy)^{2} = x-iy$
$x^{2} + (iy)^{2} +2ixy= x-iy$
$x^{2} -y^{2} +2ixy = x-iy$
Threfore , $x^{2} - y^{2} =x$---(1) and $\;2xy=y$----(2)
From (2) , we have y=0 or $\;x=-\large\frac{1}{2}$
When y=0 , from (1) , we get $\;x^{2}-x=0\;$ i.e $\;x=0\;$ or $\;x=1$
When $\;x=-\large\frac{1}{2}\;$ , from (1) , we get $\;y^{2}=\large\frac{1}{4}+\large\frac{1}{2}\;$ i.e $\;y^{2}=\large\frac{3}{4}\;$ so $\;y=\pm\large\frac{\sqrt{3}}{2}$
Hence the solutions of the given equations are
$=-\large\frac{1}{2} \pm \large\frac{\sqrt{3}}{2}$