If omega is a complex cube root of unity whos | vedclass

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(C) Given $|z - \omega| = |z + \omega|$.This represents the locus of $z$ as the perpendicular bisector of the line segment joining $\omega$ and $-\ome

Vedclass · Accepted Answer

$\frac{\pi}{6}$

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(C) Given $|z - \omega| = |z + \omega|$.This represents the locus of $z$ as the perpendicular bisector of the line segment joining $\omega$ and $-\omega$.Since $\omega = e^{i2\pi/3} = -\frac{1}{2} + i\frac{\sqrt{3}}{2}$,the points are $\omega$ and $-\omega = \frac{1}{2} - i\frac{\sqrt{3}}{2}$.The line segment joining $\omega$ and $-\omega$ passes through the origin and makes an angle of $120^{\circ}$ (or $2\pi/3$) with the positive $x$-axis.The perpendicular bisector of this line segment passes through the origin and makes an angle of $120^{\circ} + 90^{\circ} = 210^{\circ}$ or $120^{\circ} - 90^{\circ} = 30^{\circ}$ (i.e.,$\frac{\pi}{6}$) with the positive $x$-axis.Thus,$arg(z) = \frac{\pi}{6}$ or $arg(z) = \frac{7\pi}{6}$.Comparing with the options,the correct value is $\frac{\pi}{6}$.

If $\omega$ is a complex cube root of unity whose imaginary part is positive and $|z - \omega| = |z + \omega|$,then $arg(z)$ can be-

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