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प्रश्न
The real value of α for which the expression `(1 - i sin alpha)/(1 + 2i sin alpha)` is purely real is ______.
पर्याय
`(n + 1) pi/2`
`(2n + 1) pi/2`
nπ
None of these, where n ∈N
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उत्तर
The real value of α for which the expression `(1 - i sin alpha)/(1 + 2i sin alpha)` is purely real is nπ.
Explanation:
Let z = `(1 - i sin alpha)/(1 + 2i sin alpha)`
= `((1 - i sin alpha)(1 - 2i sin alpha))/((1 + 2i sin alpha)(1 - 2i sin alpha))`
= `(1 - 2i sin alpha - i sin alpha + 2i^2 sin^2 alpha)/((1)^2 - (2i sin alpha)^2`
= `(1 - 3i sin alpha - 2 sin^2 alpha)/(1 - 4i^2 sin^2 alpha)`
= `((1 - 2 sin^2 alpha) - 3i sin alpha)/(1 + 4 sin^2 alpha)`
= `(1 - 2 sin^2 alpha)/(1 + 4 sin^2 alpha) - (3sin alpha)/(1 + 4 sin^2 alpha) .i`
Since, z is purely real.
Then `(-3 sin alpha)/(1 + 4 sin^2 alpha)` = 0
⇒ sinα = 0
So, α = nπ, n ∈ N.
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