Question

The real value of α for which the expression `(1 - i sin alpha)/(1 + 2i sin alpha)` is purely real is ______.

Options

• `(n + 1)  pi/2`

• `(2n + 1)  pi/2`

• nπ

• None of these, where n ∈N

Answer 1

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.