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If $z=\cos \Theta + i \sin \Theta,$ then:

Option 1)
$z^{n}+ \frac{1}{z^{n}}=2 \cos^{n}\Theta$

Option 2)
$z^{n}+ \frac{1}{z^{n}}=2^{n} \cos{n}\Theta$

Option 3)
$z^{n}- \frac{1}{z^{n}}=2^{n} i \sin(n\Theta)$

Option 4)
$z^{n}- \frac{1}{z^{n}}=(2 i )^{n} \sin(n\Theta)$

Answers (1)

As we learnt in

Euler's Form of a Complex number -

$z=re^{i\theta}$

- wherein

r denotes modulus of z and $\theta$ denotes argument of z.

$z=cos \theta+isin \theta=e^{i \theta}\\*\\* z^{n}+ \frac{1}{ z^{n}}=e^{in \theta}+e^{-in \theta}\\*\\cosn \theta+isinn \theta+cosn \theta-isin n \theta=2cosn\theta$

Option 1)

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

Correct

Option 2)

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

Incorrect

Option 3)

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

Incorrect

Option 4)

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

Incorrect

Posted by

Vakul

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