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Slope Form of Tangent: Ellipse

If the comple...

Question

If the complex number $z_{1}, z_{2}$ the origin form an equilateral triangle then $z_{1}^{2} + z_{2}^{2}$ =

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Solution

The correct option is A
Let OA, OB be the sides of an equilateral and let OA, OB represent the complex numbers or vectors $z_{1}, z_{2}$ respectively.

From the equilateral $Δ O A B, ¯ A B = z_{2} - z_{1} ∴ a r g (\frac{z_{2} - z_{1}}{z_{1}}) = a r g (z_{2} - z_{1}) - a r g z_{2} = \frac{π}{3} a n d a r g (\frac{z_{2}}{z_{1}}) = a r g (z_{2}) - a r g (z_{1}) = \frac{π}{3} A l s o ∣ ∣ \frac{z_{2} - z_{1}}{z_{2}} ∣ ∣ = 1 = ∣ ∣ \frac{z_{2}}{z_{1}} ∣ ∣, s i n c e t r i a n g l e i s e q u i l a t e r a l .$
$T h u s t h e v e c t o r s \frac{z_{2} - z_{1}}{z_{2}} a n d \frac{z_{2}}{z_{1}} h a v e s a m e m o d u l u s a n d s a m e a r g u m e n t, w h i c h i m p l i e s t h a t t h e v e c t o r s a r e e q u a l, t h a t i s \frac{z_{2} - z_{1}}{z_{2}} = \frac{z_{2}}{z_{1}} \Rightarrow z_{1} z_{2} - z_{1}^{2} = z_{2}^{2} \Rightarrow z_{1}^{2} + z_{2}^{2} = z_{1} z_{2}$
Note : Students should remember this question as a formula.

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