Example 15 Find real θ such that3+2isin0is purely real.

1.	Example 15 Find real θ such that3+2isin0is purely real.
Answer» Rationalize the denominator = [(3 + 2i sinθ)(1 + 2i sinθ)] / [(1 - 2i sinθ)(1 + 2i sinθ)] = (3 + 6i sinθ + 2i sinθ - 4sin^2θ) / (1 + 4sin2θ) = (3 - 4sin2θ /1 + 4sin^2θ) (8i sinθ /1 + 4sin^2θ) Given: the complex numbers are real. Therefore 8i sinθ/1 + 4sin^2θ=0 i.e; sinθ = 0 Thus θ = nπ, n∈ Z

Answer»

Rationalize the denominator

= [(3 + 2i sinθ)(1 + 2i sinθ)] / [(1 - 2i sinθ)(1 + 2i sinθ)]

= (3 + 6i sinθ + 2i sinθ - 4sin^2θ) / (1 + 4sin2θ)

= (3 - 4sin2θ /1 + 4sin^2θ) (8i sinθ /1 + 4sin^2θ)

Given: the complex numbers are real. Therefore

8i sinθ/1 + 4sin^2θ=0

i.e; sinθ = 0

Thus θ = nπ, n∈ Z

Discussion