If the expression 1−isinα1+2isinα is purely real, which of following is/are correct?
A
Re(z)=−15
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B
Re(z)=1
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C
α=nπ,n∈Z
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D
α=nπ2,n∈Z
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Solution
The correct options are BRe(z)=1 Cα=nπ,n∈Z 1−isinα1+2isinα =1−isinα1+2isinα.1−2isinα1−2isinα =(1−2sin2α)−3isinα1+4sin2α =1−2sin2α1+4sin2α−3sinα1+4sin2αi Which is purely real if 3sinα1+4sin2α=0 ⇒sinα=0 ⇒α=nπ,n∈Z