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If z= { } +i { } , then

Question

If $z = cos θ + i sin θ$ , then

z^{n} + \frac{1}{z ^{n}} = 2 cos n θ

z^{n} + \frac{1}{z ^{n}} = 2^{n} cos n θ

z^{n} - \frac{1}{z ^{n}} = 2 i sin n θ

z^{n} - \frac{1}{z ^{n}} = (2 i)^{n} sin n θ

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Chloe Discussed

z = cos θ + i sin θ

, then

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Text solutionVerified

We have

\frac{1}{z} = \frac{1}{cos θ + i sin θ} = cos θ - i sin θ

∴ z^{n} = (cos θ + i sin θ)^{n} = cos n θ + i sin n θ

and

\frac{1}{z ^{n}} = (cos θ - i sin θ)^{n} = cos n θ - i sin n θ

Hence

z^{n} + \frac{1}{z ^{n}} = 2 cos n θ

and

z^{n} - \frac{1}{z ^{n}} = 2 i sin n θ

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