Let $z_{1}, z_{2}$ and $z_{3}$ represent the vertices A, B and C of the triangle $A B C$ in the Argand plane, such that $| z_{1} | = | z_{2} | = | z_{3} | = 5$ , then the value of $z_{1} sin 2A + z_{2} sin 2 B + z_{3} sin 2C$ is

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Solution

$| z_{1} | = | z_{2} | = | z_{3} | = 5$
$\Rightarrow | z | = 5$ is the circumcircle of triangle ABC
$\Rightarrow ∠ A O B = 2 C, ∠ B O C = 2 A$ and $∠ C O A = 2 B$
We have,
$\frac{z_{2}}{z_{1}} = \frac{O B}{O A} e^{i 2 C} = e^{i 2 C}$
Similarly,
$\frac{z_{3}}{z_{1}} = \frac{O C}{O A} e^{- i 2 B} = e^{- i 2 B}$
Now,
$z_{1} sin 2A + z_{2} sin 2B + z_{3} sin 2C$
$= z_{1} (sin 2A + \frac{z_{2}}{z_{1}} sin 2B + \frac{z_{3}}{z_{1}} sin 2C)$
$= z_{1} (sin 2A + e^{i 2 C} sin 2B + e^{- 2 i B} sin 2C)$
$= z_{1} (sin 2A + sin 2B cos 2C + i sin 2B sin 2C + sin 2C cos 2B - i sin 2C sin 2B)$
$= z_{1} (sin 2A + sin (2B + 2C))$
$= z_{1} (sin 2A + sin (2 π - 2 A)) = 0$

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