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Area and Volume Expansion

If z 1 , z ...

Question

If $z_{1}, z_{2}, z_{3}$ are the vertices of an equilateral triangle in the complex plane and $z_{0}$ is the centroid, then

A

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

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B

{(z_{1} - z_{2})}^{2} + {(z_{2} - z_{3})}^{3} + {(z_{3} - z_{1})}^{2} = 0

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C

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

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D

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

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Solution

The correct options are
A $\frac{1}{z_{1} - z_{2}} + \frac{1}{z_{2} - z_{3}} + \frac{1}{z_{3} - z_{1}} = 0$
B ${z_{1}}^{2} + {z_{2}}^{2} + {z_{3}}^{2} = 3 {z_{0}}^{2}$
C ${(z_{1} - z_{2})}^{2} + {(z_{2} - z_{3})}^{3} + {(z_{3} - z_{1})}^{2} = 0$
D ${z_{1}}^{2} + {z_{2}}^{2} + {z_{3}}^{2} = z_{1} z_{2} + z_{2} z_{3} + z_{3} z_{1}$
Let $A, B, C$ be the vertices of the equilateral triangle represented by the complex number $z_{1}, z_{2}, z_{3}$ , respectively.
Then $A B = B C = A C$ and $∠ A = ∠ B = ∠ C = \frac{π}{3}$
$∴ \frac{z_{3} - z_{1}}{z_{2} - z_{1}} = e^{i π / 3} = \frac{z_{1} - z_{2}}{z_{3} - z_{2}} = \frac{z_{2} - z_{3}}{z_{1} - z_{3}}$ ...(1)
$\Rightarrow (z_{3} - z_{1}) (z_{3} - z_{2}) = {- (z_{1} - z_{2})}^{2}$
$\Rightarrow {z_{1}}^{2} + {z_{2}}^{2} + {z_{3}}^{2} = z_{1} z_{2} + z_{2} z_{3} + z_{3} z_{1}$ ...(2)
$\Rightarrow \frac{1}{2} [{(z_{1} - z_{2})}^{2} + {(z_{2} - z_{3})}^{3} + {(z_{3} - z_{1})}^{2}] = 0$ ...(3)
Now, $3 z_{0} = z_{1} + z_{2} + z_{3}$ $[z_{0}$ is controid $]$
$∴ {z_{1}}^{2} + {z_{2}}^{2} + {z_{3}}^{2} + 2 (z_{1} z_{2} + z_{2} z_{3} + z_{3} z_{1}) = 9 {z_{0}}^{2}$
$\Rightarrow {z_{1}}^{2} + {z_{2}}^{2} + {z_{3}}^{2} = 3 {z_{0}}^{2}$ ...(4)
From (1), we also have
$(z_{3} - z_{1}) (z_{3} - z_{2}) + (z_{1} - z_{2}) (z_{1} - z_{2}) (z_{1} - z_{3} + z_{3} - z_{2}) = 0 \Rightarrow (z_{2} - z_{1}) (z_{3} - z_{1}) + (z_{1} - z_{2}) (z_{3} - z_{2}) + (z_{1} - z_{2}) (z_{1} - z_{3}) = 0$
Dividing by $(z_{1} - z_{2}) (z_{2} - z_{3}) (z_{3} - z_{1})$ , we get
$\frac{1}{z_{1} - z_{2}} + \frac{1}{z_{2} - z_{3}} + \frac{1}{z_{3} - z_{1}} = 0$

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Similar questions

z_{1}, z_{2}, z_{3} \in

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

z_{1} + z_{2} + z_{3} \neq 0

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

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

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

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

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

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 equilateral triangle with centroid

z_{0}

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

z_{1}, z_{2}

z_{3}

be three complex numbers such that

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

\frac{z_{1}^{2}}{z_{2} z_{3}} + \frac{z_{2}^{2}}{z_{3} z_{1}} + \frac{z_{3}^{2}}{z_{1} z_{2}} + 1 = 0.

Then the sum of all possible values of

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

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