First slide

Geometry of complex numbers

Question

$From a point P representing complex number z_{1} on the curve (| z | = 2) two tangents are drawn form$
$P to the curve | z | = 1 meets at A (z_{2}) and B (z_{3}) then$

Difficult

A

$centroid of ΔABP will be on curve | z | = 1$
B

$\arg (\frac{z_{2}}{z_{3}}) = \pm \frac{2 π}{3}$
C

$(\frac{4}{\bar{z_{1}}} + \frac{1}{\bar{z_{2}}} + \frac{1}{\bar{z_{3}}}) (\frac{4}{z_{1}} + \frac{1}{z_{2}} + \frac{1}{z_{3}}) = 9$
D

$orthocenter and circumcentre of ΔABP will coincide.$

Solution

$\begin{array}{l} O A = O B = 1; O P = 2 \\ \sin ∠ O P A = \frac{1}{2} ∠ O P A = \frac{π}{6} ∠ A P B = \frac{π}{3} \end{array}$
$OM ⊥ AB so △ APB is equilateral. so (D) is true.$
$OM = 1 / 2; MN = 1 / 2 PN = 1 so$
$centroid lies on | z | = 1 - - - - (A)$
$\begin{array}{l} |\frac{z_{1} + z_{2} + z_{3}}{3}| = 1 \\ {|z_{1} + z_{2} + z_{3}|}^{2} = 9 \\ (z_{1} + z_{2} + z_{3}) (\bar{z_{1}} + \bar{z_{2}} + \bar{z_{3}}) = 9 \\ S i n c e z_{1} \bar{z_{1}} = 4, z_{2} \bar{z_{2}} = 1 a n d z_{3} \bar{z_{3}} = 1 \\ ⇒ (\frac{4}{\bar{z_{1}}} + \frac{1}{\bar{z_{2}}} + \frac{1}{\bar{z_{3}}}) (\frac{4}{z_{1}} + \frac{1}{z_{2}} + \frac{1}{z_{3}}) = 9 - - - - (B) \\ ∠ AOB = π - \frac{π}{3} = \frac{2 π}{3} - - - (C) \end{array}$

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