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Properties of nth Root of a Complex Number

If z1, z2, z3...

Question

If $z_{1}, z_{2}, z_{3} . . . . . n_{n}$ are n^th, roots of unity, then for k = 1, 2, ....., n

A

| z_{k} | = k | z_{k + 1} |

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B

| z_{k + 1} | = k | z_{k} |

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C

| z_{k + 1} | = | z_{k} | + | z_{k + 1} |

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D

| z_{k} | = | z_{k + 1} |

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Solution

The correct option is D $| z_{k} | = | z_{k + 1} |$
The $n^{t h}$ roots of unity are given by $z_{k} = e^{\frac{i 2 π (k - 1)}{n}}, (k = 1, 2, . . . . ., n)$

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Similar questions

z_{1}, z_{2}, z_{3} . . . . . n_{n}

are n^th, roots of unity, then for k = 1, 2, ....., n

z_{1}, z_{2}

are two complex numbers and

ω^{k}, k = 0, 1, . . ., n - 1

are the nth roots of unity, then

n - 1 \sum k = 0 {∣ ∣ z_{1} + z_{2} ω^{k} ∣ ∣}^{2}

1, z_{1}, z_{2}, z_{3}, . . . . ., z_{n - 1}

are the nth roots of unity, then the value of

\frac{1}{(3 - z_{1})} + \frac{1}{(3 - z_{2})} + . . . . . + \frac{1}{(3 - z_{n - 1})}

1, z_{1}, z_{2}, z_{3}, . . ., z_{n - 1}

n

th roots of unity, then show that

(1 - z_{1}) (1 - z_{2}) . . . (1 - z_{n - 1}) = n

1, ω, ω^{2}, \dots \dots ω^{n - 1}

n^{th}

roots of unity and

z_{1}

z_{2}

are any two complex numbers, then

n - 1 \sum k = 0 | z_{1} + ω^{k} z_{2} |^{2}

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