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Given $z^2+z|z|+|z^2|=0$ the locus of z is

$\begin{array}{1 1}(A)\;\text{A straight line}&(B)\;\text{A circle}\\(C)\;\text{A pair of straight line}&(D)\;\text{None of these}\end{array} $

1 Answer

Let $z=r(\cos \theta+i\sin \theta)=re^{i\theta}$
$z^2=r^2e^{i2\theta}$
$z^2+z|z|+|z^2|=0$
$\Rightarrow r^2e^{2i\theta}+r^2e^{i\theta}+r^2=0$
$\Rightarrow (e^{i\theta})^2+e^{i\theta}+1=0$
$\Rightarrow e^{i\theta}=\large\frac{-1\pm \sqrt {3i}}{2}$
$\Rightarrow \cos\theta=-\large\frac{1}{2}$$\sin \theta=\pm \large\frac{\sqrt 3}{2}$
$r$ is variable
Thus locus is a pair of straight lines.
Hence (C) is the correct answer.
answered Apr 10, 2014 by sreemathi.v
 

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