If $z_{1}$ , $z_{2}$ , $z_{3}$ are vertices of an equilateral triangle then the value of $(\frac{1}{z_{1} - z_{2}} + \frac{1}{z_{2} - z_{3}} + \frac{1}{z_{3} - z_{1}})$ is:

0

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

1

No worries! We‘ve got your back. Try BYJU‘S free classes today!

2

No worries! We‘ve got your back. Try BYJU‘S free classes today!

3

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is A $0$
Solving,
$\frac{1}{z_{1} - z_{2}} + \frac{1}{z_{2} - z_{3}} + \frac{1}{z_{3} - z_{1}}$
Numerator $= (z_{2} - z_{3}) (z_{3} - z_{1}) + (z_{1} - z_{2}) (z_{3} - z_{1}) + (z_{1} - z_{2}) (z_{2} - z_{3})$
$= z_{2} z_{3} - z_{1} z_{2} - z {_{3}}^{2} + z_{3} z_{1} + z_{1} z_{3} - z {_{1}}^{2} - z_{2} z_{3} + z_{1} z_{2}$
$+ z_{1} z_{2} - z_{1} z_{3} - z {_{2}}^{2} + z_{2} z_{3}$
$= - (z {_{1}}^{2} + z {_{2}}^{2} + z {_{3}}^{2}) + (z_{1} z_{3} + z_{2} z_{3} + z_{1} z_{2})$
$= 0$ (for an equilateral $Δ$ )

Suggest Corrections

Similar questions

| z_{1} | = | z_{2} | = | z_{3} |

z_{1}, z_{2}, z_{3}

z_{1} + z_{2} + z_{3}

z_{1}, z_{2}, z_{3}

| z_{1} | = | z_{2} | = | z_{3} |

z_{1}, z_{2}, z_{3}

ω^{3} = 1, ω \neq 1

z + ω z_{2} + ω^{2} z_{3} = 0

z_{1}, z_{2}, z_{3}

z_{3} - z_{1} = (z_{2} - z_{1}) e^{- 1^{π} / 3}

z_{1}, z_{2}, z_{3}

z_{1}, z_{2}, z_{3}

\frac{1}{z_{1} - z_{2}} + \frac{1}{z_{2} - z_{3}} + \frac{1}{z_{3} - z_{1}} = 0

z_{1}, z_{2}, z_{3}

\frac{1}{z_{2} - z_{3}} + \frac{1}{z_{3} - z_{1}} + \frac{1}{z_{1} - z_{2}} = 0

Join BYJU'S Learning Program