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If $A (z_{1}) B (z_{2}) C (z_{3})$ are the vertices of an equilateral triangle ABC then the
value of Arg $(\frac{z_{2} + z_{3} - 2 z_{1}}{z_{3} - z_{2}}) =______$

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$\frac{π}{3}$

$\frac{π}{4}$

$\frac{π}{6}$

$\frac{π}{2}$

answer is B.

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Detailed Solution

$A r g (2 (\frac{\frac{z_{2} + z_{3}}{2} - z_{1}}{z_{3} - z_{2}}))$

$= A r g (\frac{\frac{z_{2} + z_{3}}{2} - z_{1}}{z_{3} - z_{2}}) = \frac{π}{2}$

$\frac{z_{3} + z_{2}}{2}$ in the mid point of side BC clearly line joining A and mid point of BC with be $⊥$ er to side BC

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