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Inequality of 2 Complex Numbers

If z_1,z_2,z_...

Question

If $z_{1}, z_{2}, z_{3}$ are the vertices of a triangle in argand plane such that $| z_{1} - z_{2} | = | z_{1} - z_{3} |,$ then $arg (\frac{2 z_{1} - z_{2} - z_{3}}{z_{3} - z_{2}})$ is

\pm \frac{π}{3}

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\pm \frac{π}{2}

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\pm \frac{π}{6}

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Solution

The correct option is C $\pm \frac{π}{2}$

$\frac{z_{1} - z_{2}}{z_{3} - z_{2}} = \frac{| z_{1} - z_{2} |}{| z_{3} - z_{2} |} e^{i θ}$
$\Rightarrow \frac{z_{1} - z_{3}}{z_{2} - z_{3}} = \frac{| z_{1} - z_{3} |}{| z_{2} - z_{3} |} e^{- i θ}$
$\Rightarrow \frac{z_{1} - z_{2}}{z_{3} - z_{2}} + \frac{z_{1} - z_{3}}{z_{3} - z_{2}} =$ Purely imaginary number
So, $arg (\frac{2 z_{1} - z_{2} - z_{3}}{z_{3} - z_{2}}) = \pm \frac{π}{2}$

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$| Z_{1} | = | Z_{2} | = | Z_{3} | = | \frac{1}{Z_{1}} + \frac{1}{Z_{2}} + \frac{1}{Z_{3}} | = 1$ then $| Z_{1} + | Z_{2} + Z_{3} |$ is

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