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

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

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Let $z=x+iy$ then $\overline z=x-iy$
$z^2+\overline z=x^2-y^2+2xyi+x-yi=0$
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)
case (ii)
If $x=\large\frac{1}{2},$ then
$\Rightarrow\:y=\pm\large\frac{\sqrt 3}{2}$
$i.e.,z=\large\frac{1}{2}\pm\frac{\sqrt 3}{2}i$
$\therefore$ There are four values for $z$


answered Jul 2, 2013 by rvidyagovindarajan_1
edited Jul 2, 2013 by rvidyagovindarajan_1

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