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Question

If (cosθ+isinθ)(cos2θ+isin2θ)

(cosnθ+isinnθ)=1, then the value of θ is


A

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B

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C

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Solution

The correct option is C


We have (cosθ+isinθ)(cos2θ+isin2θ)

..........(cosnθ+isinnθ)=1

(cos(θ+2θ+3θ+.......nθ)+isin(θ+2θ+3θ+.......nθ)=1

cos(n(n+1)2θ) + isin(n(n+1)2θ)=1

cos(n(n+1)2θ)=1 and sin(n(n+1)2θ)=0

n(n+1)2θ = 2mπ θ= 4mπn(n+1), Where m I


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