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Question

If tan(α+iβ)=eiθ ; where α,βR, θ(2n+1)π2,nZ and i=1, then

A
α is an odd multiple of π2
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B
α is an even multiple of π2
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C
α is an odd multiple of π4
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D
α is an even multiple of π4
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Solution

The correct option is C α is an odd multiple of π4
tan(α+iβ)=eiθ
tan(α+iβ)=cosθ+isinθ
Taking complex conjugate, we get
tan(αiβ)=cosθisinθ

tan2α=tan[(α+iβ)+(αiβ)]

tan2α=tan(α+iβ)+tan(αiβ)1tan(α+iβ)tan(αiβ)

1tan2α=1tan(α+iβ)tan(αiβ)tan(α+iβ)+tan(αiβ)

cot2α=1(cos2θ+sin2θ)2cosθ

cot2α=0 [θ(2n+1)π2]

2α=(2m+1)π2,mZ

α=(2m+1)π4,mZ

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