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# The number of solutions of the equation $z^2+\overline {z}=0$ is

$\begin{array}{1 1}(A) \;2 \\(B)\;4 \\(C)\;6 \\(D)\;8 \end{array}$

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

edited Jul 2, 2013