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Complex Numbers

Solutions of ...

Question

Solutions of the equation $z^{7} - 1 = 0$ are given by

z = - 1, z = cos \frac{2 k π}{7} + i sin \frac{2 k π}{7}, k = 0, 1, 2, 3, 4, 5

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z = 1 a n d z = cos \frac{2 k π}{7} + i sin \frac{2 k π}{7}, k = 1, 2, 3, 4, 5, 6

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z = - 1, z = cos \frac{k π}{7} + i sin \frac{k π}{7}, k = 0, 1, 2, 3, 4, 5

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z = 1 a n d z = cos \frac{k π}{7} + i sin \frac{k π}{7}, k = 0, 1, 2, 3, 4, 5

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Solution

The correct option is B $z = 1 a n d z = cos \frac{2 k π}{7} + i sin \frac{2 k π}{7}, k = 1, 2, 3, 4, 5, 6$
$z^{7} = 1 \Rightarrow z = 1^{\frac{1}{7}} = {(cos 0 + i sin 0)}^{\frac{1}{7}}$
$\Rightarrow z = cos \frac{2 k π}{7} + i sin \frac{2 k π}{7}$ ...{ De Moivre's Theorem}
Where $k = 0, 1, 2, 3, 4, 5, 6$

For $k = 0$
$z = 1$
And $z = cos \frac{2 k π}{7} + i sin \frac{2 k π}{7}$ for $k = 1, 2, 3, 4, 5, 6$
Ans:B

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