Question

The real value of α for which the expression ${\displaystyle \frac{1-i\sin \alpha }{1+2i\sin \alpha }}$ is purely real is ______.

विकल्प

• ${\displaystyle (n+1) \frac{\pi }{2}}$

• ${\displaystyle (2n+1) \frac{\pi }{2}}$

• nπ

• None of these, where n ∈N

Answer 1

The real value of α for which the expression ${\displaystyle \frac{1-i\sin \alpha }{1+2i\sin \alpha }}$ is purely real is nπ.

Explanation:

Let z = ${\displaystyle \frac{1-i\sin \alpha }{1+2i\sin \alpha }}$

= ${\displaystyle \frac{(1-i\sin \alpha )(1-2i\sin \alpha )}{(1+2i\sin \alpha )(1-2i\sin \alpha )}}$

= ${\displaystyle \frac{1-2i\sin \alpha -i\sin \alpha +2i^2{\sin }^2\alpha }{{\left(1\right)}^2-{\left(2i\sin \alpha \right)}^2}}$

= ${\displaystyle \frac{1-3i\sin \alpha -2{\sin }^2\alpha }{1-4i^2{\sin }^2\alpha }}$

= ${\displaystyle \frac{(1-2{\sin }^2\alpha )-3i\sin \alpha }{1+4{\sin }^2\alpha }}$

= ${\displaystyle \frac{1-2{\sin }^2\alpha }{1+4{\sin }^2\alpha }-\frac{3\sin \alpha }{1+4{\sin }^2\alpha }.i}$

Since, z is purely real.

Then ${\displaystyle \frac{-3\sin \alpha }{1+4{\sin }^2\alpha }}$ = 0

⇒ sinα = 0

So, α = nπ, n ∈ N.