A, B, C are the points representing the complex numbers $z_{1}, z_{2}, z_{3}$ respectively on the
complex plane and the circumcentre of the triangle ABC lies at the origin. If the altitude AD of the triangle ABC meets the circumcircle again at P, then P represents the complex number

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$- \frac{{\bar{z}}_{1} z_{2}}{{\bar{z}}_{3}}$

$- {\bar{z}}_{1} z_{2} z_{3}$

$- \frac{z_{2} z_{3}}{z_{1}}$

$- \frac{{\bar{z}}_{1} z_{3}}{{\bar{z}}_{2}}$

answer is B, C, D.

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Detailed Solution

$A P ⊥ B C \Rightarrow s u m o f c o m p l e x s l o p e s = z e r o$

$\Rightarrow \frac{z_{4} - z_{1}}{{\bar{z}}_{4} - {\bar{z}}_{1}} + \frac{z_{2} - z_{3}}{{\bar{z}}_{2} - {\bar{z}}_{3}} = 0 u s e \bar{z} = \frac{R_{2}}{z} \Rightarrow z_{4} = - \frac{z_{2} z_{3}}{z_{1}} = - \frac{{\bar{z}}_{1} z_{2}}{{\bar{z}}_{3}} = \frac{- {\bar{z}}_{1} z_{3}}{{\bar{z}}_{2}}$