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Area Method to Find Condition for Co-Linearity

If |z1|=|z2|...

Question

If $| z_{1} | = | z_{2} | = | z_{3} | = 1$ and $z_{1}, z_{2}, z_{3}$ are represented by the vertices of an equilaterial triangle then

A

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

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B

z_{1} . z_{2} . z_{3} = 1

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C

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

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D

none of these

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Solution

The correct options are
A $z_{1} + z_{2} + z_{3} = 0$
C $z_{1} . z_{2} . z_{3} = 1$
Consider the cube roots of unity, which form an equilateral triangle of unit length.
Hence
$w, w^{2}, 1$ are the vertices of the triangle.
Now
$1 + w + w^{2} = 0$
Hence
$z_{1} + z_{2} + z_{3} = 0$
and
$1. w . w^{2} = w^{3} = 1$
Hence
$z_{1} . z_{2} . z_{3} = 1$

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Similar questions

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

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

are represented by the vertices of an equilateral triangle. Then

| Z_{1} | + | Z_{2} | + | Z_{3} | = | Z_{1} + Z_{2} + Z_{3} |

\frac{Z_{1} Z_{2}}{Z_{3}^{2}} + \frac{Z_{2} Z_{3}}{Z_{2}^{2}} + \frac{Z_{3} Z_{1}}{Z_{2}^{2}}

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

represent the vertices of an equilateral triangle such that

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

Q. Assertion :Let

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

be three complex numbers such that

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

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

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

will represent vertices of an equilateral triangle on the complex plane. Reason:

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

represent vertices of an equilateral triangle if

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

.

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

are the vertices of an equilaterial triangle circumscribing the circle

| z | = 1

z_{1} = 1 + \sqrt{3} i

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

are in the anticlockwise sense then

z_{2}

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Area Method to Find Condition for Co-Linearity

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