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Question

The real value of θ for which the expression, 1+icosθ12icosθ is real number is

A
nπ±π2
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B
nππ2
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C
nπ+π2
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D
2nπ±π2
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Solution

The correct option is A nπ±π2
Let z=1+icosθ12icosθ
=(1+icosθ)(1+2icosθ)(12icosθ)(1+2icosθ)
=12cos2θ+i3cosθ1+4cos2θ
=12cos2θ1+4cos2θ+i(3cosθ1+4cos2θ)
If the above complex number is purely real, then
Im(z)=0
3cosθ=0
θ=(2n±1)π2, where nϵN

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