The complex number $z_{1}, z_{2}$ and the origin form an equilateral triangle only if $\frac{z_{1}}{z_{2}} + \frac{z_{2}}{z_{1}} = 1$ .

True

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False

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Solution

The correct option is A True
$L e t A B C b e a e q u i l a t e r a l t r i a n g l e$
$A (0, 0), B (z_{1}), C (z_{2})$
$A B = | z_{1} |, A C = | z_{2} |, B C = | z_{1} - z_{2} |$
$sin c e A B C i s a n e q u i l a t e r a l t r i a n g l e$
$L e t | z_{1} | = | z_{2} | = | z_{1} - z_{2} | = k$
$c a s e 1$
$| z_{1} | = k$
$s q u a r i n g b o t h s i d e s$
${| z_{1} |}^{2} = k^{2}$
$\Rightarrow (z_{1}) (_{1}) = k^{2}$
$\Rightarrow_{1} = \frac{k^{2}}{z_{1}}$
$\Rightarrow -_{1} = \frac{- k^{2}}{z_{1}} . . . . . . . . . (i)$
$c a s e 2$
$| z_{2} | = k$
$s q u a r i n g b o t h s i d e s$
${| z_{2} |}^{2} = k^{2}$
$\Rightarrow (z_{2}) (_{2}) = k^{2}$
$\Rightarrow_{2} = \frac{k^{2}}{z_{2}}$
$\Rightarrow_{2} = \frac{k^{2}}{z_{2}} . . . . . . . . . (i i)$
$c a s e 3$
$| z_{1} - z_{2} | = k$
$s q u a r i n g b o t h s i d e s$
${| z_{1} - z_{2} |}^{2} = k^{2}$
$\Rightarrow (z_{1} - z_{2}) (¯ ¯¯¯¯¯¯¯¯¯¯¯¯¯¯ ¯ z_{1} - z_{2}) = k^{2}$
$\Rightarrow ¯ ¯¯¯¯¯¯¯¯¯¯¯¯¯¯ ¯ z_{1} - z_{2} = \frac{k^{2}}{z_{1} - z_{2}}$
$\Rightarrow ¯ ¯¯¯¯¯¯¯¯¯¯¯¯¯¯ ¯ z_{1} - z_{2} = \frac{k^{2}}{z_{1} - z_{2}} . . . . . . . . . (i i i)$
$a d d i n g (i) (i i) (i i i)$
$0 = k^{2} [\frac{- 1}{z_{1}} + \frac{1}{z_{2}} + \frac{1}{z_{1} - z_{2}}]$
$\Rightarrow 0 = \frac{- 1}{z_{1}} + \frac{1}{z_{2}} + \frac{1}{z_{1} - z_{2}}$
$\Rightarrow 0 = \frac{- z_{2} (z_{1} - z_{2}) + z_{1} (z_{1} - z_{2}) + z_{1} z_{2}}{z_{1} z_{2} (z_{1} - z_{2})}$
$\Rightarrow 0 = - z_{1} z_{2} + z_{2}^{2} + z_{1}^{2} - z_{1} z_{2} + z_{1} z_{2}$
$\Rightarrow 0 = - z_{1} z_{2} + z_{2}^{2} + z_{1}^{2}$
$\Rightarrow z_{1} z_{2} = z_{2}^{2} + z_{1}^{2}$
$\Rightarrow 1 = \frac{z_{2}}{z_{1}} + \frac{z_{1}}{z_{2}}$
$H e n c e t h e g i v e n s t a t e m e n t i s t r u e$

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