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## QuestionMathsClass 11

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

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 $$\displaystyle{ z }^{ n }-\frac { 1 }{ { z }^{ n } } =2i\sin { n\theta }$$