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Q.

1+sin θ+i cos θn=

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a

2n.cosnπ6-θ2.cisnπ6-θ2

b

2n.cosnπ3-θ2.cisnπ3-θ2

c

2n.cosnπ4-θ2.cisnπ4-θ2

d

2n.cosnπ5-θ2.cisnπ5-θ2

answer is B.

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Detailed Solution

  (1+sin θ+i cos θ)n   1+cos π2-θ+i sin π2-θn   2cos2π4-θ2+i2sin π4-θ2cosπ4-θ2n  =2cosπ4-θ2n cosπ4-θ2+i sinπ4-θ2n =2n cosn π4-θ2 cos nπ4-θ2+i sin nπ4-θ2 =2n cosn π4-θ2 cis nπ4-θ2

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