If \(z=\cos { \theta } +i\sin { \theta } \), then

(A) \(\displaystyle{ z }^{ n }+\frac { 1 }{ { z }^{ n } } =2\cos { n\theta } \)

(B) \(\displaystyle{ z }^{ n }+\frac { 1 }{ { z }^{ n } } ={ 2 }^{ n }\cos { n\theta } \)

(C) \(\displaystyle{ z }^{ n }-\frac { 1 }{ { z }^{ n } } ={ 2 }i\sin { n\theta } \)

(D) \(\displaystyle{ z }^{ n }-\frac { 1 }{ { z }^{ n } } ={ \left( 2i \right) }^{ n }\sin { n\theta } \)

(A) \(\displaystyle{ z }^{ n }+\frac { 1 }{ { z }^{ n } } =2\cos { n\theta } \)

(B) \(\displaystyle{ z }^{ n }+\frac { 1 }{ { z }^{ n } } ={ 2 }^{ n }\cos { n\theta } \)

(C) \(\displaystyle{ z }^{ n }-\frac { 1 }{ { z }^{ n } } ={ 2 }i\sin { n\theta } \)

(D) \(\displaystyle{ z }^{ n }-\frac { 1 }{ { z }^{ n } } ={ \left( 2i \right) }^{ n }\sin { n\theta } \)

Answer: A C

We have \(\displaystyle\frac { 1 }{ z } =\frac { 1 }{ \cos { \theta } +i\sin { \theta } } =\cos { \theta } -i\sin { \theta } \)

We have \(\displaystyle\frac { 1 }{ z } =\frac { 1 }{ \cos { \theta } +i\sin { \theta } } =\cos { \theta } -i\sin { \theta } \)

\(\displaystyle\therefore { z }^{ n }={ \left( \cos { \theta } +i\sin { \theta } \right) }^{ n }=\cos { n\theta } +i\sin { n\theta } \)

and \(\displaystyle\frac { 1 }{ { z }^{ n } } ={ \left( \cos { \theta } -i\sin { \theta } \right) }^{ n }=\cos { n\theta } -i\sin { n\theta } \)

Hence \(\displaystyle{ z }^{ n }+\frac { 1 }{ { z }^{ n } } =2\cos { n\theta } \) and \(\displaystyle{ z }^{ n }-\frac { 1 }{ { z }^{ n } } =2i\sin { n\theta } \)

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