If $z_{1} + z_{2} + z_{3} = 0$ and $| z_{1} | = | z_{2} | = | z_{2} | = 1$ , then prove that the complex numbers $z_{1}, z_{2}, z_{3}$ are the vertices of an equilateral triangle inscribed in the unit circle.

Open in App

Solution

$\begin{matrix} z_{1} + z_{2} + z_{3} = 0 | z_{1} | = | z_{2} | = | z_{3} | = 1 \frac{z_{1} + z_{2} + z_{3}}{3} = 0 \end{matrix}$
Centro id of triangle is $(0, 0)$ .
Also, cir cum center of triangle is $(0, 0)$ as $z_{i}$ lies on circle $| z_{1} | = 1$ .
Circumcentre centro id of triangle coincide
Hence, triangle is equilateral.

Suggest Corrections

Similar questions

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}

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

z_{1} + z_{2} + z_{3} = 0

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}

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

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

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

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

Join BYJU'S Learning Program