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Modulus of a Complex Number

If Z1+Z2+Z3...

Question

If $Z_{1} + Z_{2} + Z_{3} = 0$ and $| Z_{1} | = | Z_{2} | = | Z_{3} | = 1$ , then $Z_{1}^{2} + Z_{2}^{2} + Z_{3}^{2}$ equals

A

+ 3

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B

\frac{1}{3}

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C

1

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D

0

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Solution

The correct option is D $0$

As $(z_{1} + z_{2} + z_{3}) = 0$
Therefore, $(z_{1} + z_{2} + z_{3})^{2} = 0$

$\Rightarrow z_{1}^{2} + z_{2}^{2} + z_{3}^{2} + 2 (z_{1} z_{2} + z_{2} z_{3} + z_{1} z_{3}) = 0$

But,

$\Rightarrow z_{1} z_{2} + z_{2} z_{3} + z_{3} z_{1} = z_{1} z_{2} z_{3} (\frac{1}{z_{1}} + \frac{1}{z_{2}} + \frac{1}{z_{3}})$

$\Rightarrow z_{1} z_{2} z_{3} (_{1} +_{2} +_{3}) = 0$

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

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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}

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} |

Q. Let the four roots of unity be

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

z_{4}

, respectively.
Statement 1:

z_{1}^{2} + z_{2}^{2} + z_{3}^{2} + z_{4}^{2} = 0

z_{1} + z_{2} + z_{3} + z_{4} = 0

.

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

be complex numbers such that

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} | = 21.

| z_{1} - z_{2} | = 2 \sqrt{3}, | z_{1} | = 3 \sqrt{3},

then the value of

| 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_{1} z_{3}} + \frac{z_{3}^{2}}{z_{1} z_{2}} = - 1.

Then the possible value(s) of

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

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