If {z^2} + z|z| + |z|^2 = 0,then the locus of | vedclass

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(C) Given equation: ${z^2} + z|z| + |z|^2 = 0$Divide by $|z|^2$ (assuming $z 
eq 0$):${\left( {\frac{z}{|z|}} ight)^2} + \frac{z}{|z|} + 1 = 0$Let

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$C$. $A$ pair of straight lines

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(C) Given equation: ${z^2} + z|z| + |z|^2 = 0$Divide by $|z|^2$ (assuming $z 
eq 0$):${\left( {\frac{z}{|z|}} ight)^2} + \frac{z}{|z|} + 1 = 0$Let $u = \frac{z}{|z|}$. Then $u^2 + u + 1 = 0$.The roots are $u = \omega$ or $u = \omega^2$,where $\omega = \frac{-1 + i\sqrt{3}}{2}$.Case $1$: $\frac{z}{|z|} = \omega = -\frac{1}{2} + i\frac{\sqrt{3}}{2}$$z = |z|\left( -\frac{1}{2} + i\frac{\sqrt{3}}{2} ight)$$x + iy = -\frac{1}{2}|z| + i\frac{\sqrt{3}}{2}|z|$Comparing real and imaginary parts: $x = -\frac{1}{2}|z|$ and $y = \frac{\sqrt{3}}{2}|z|$.Thus,$y = -\sqrt{3}x$,or $y + \sqrt{3}x = 0$.Case $2$: $\frac{z}{|z|} = \omega^2 = -\frac{1}{2} - i\frac{\sqrt{3}}{2}$$x + iy = -\frac{1}{2}|z| - i\frac{\sqrt{3}}{2}|z|$Comparing real and imaginary parts: $x = -\frac{1}{2}|z|$ and $y = -\frac{\sqrt{3}}{2}|z|$.Thus,$y = \sqrt{3}x$,or $y - \sqrt{3}x = 0$.Combining both,the locus is a pair of straight lines: $(y + \sqrt{3}x)(y - \sqrt{3}x) = 0$,which simplifies to $y^2 - 3x^2 = 0$.

If ${z^2} + z|z| + |z|^2 = 0$,then the locus of $z$ is

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