Question

Find real values of θ for which $\Big(\frac{4+3i\sin\theta}{1-2i\sin\theta} \Big)$ is purely real.

Answer 1

Answer:

θ = nπ, where n∈Z

Step-by-step explanation:

Hi,

Given (4 + 3i sin θ)/(1 - 2i sin θ)

Multiplying and dividing by 1 + 2i sinθ, we get

[(4 + 3i sin θ)*(1 + 2i sinθ)/(1 - 2i sin θ)*(1 + 2i sinθ)]

= [ 4 + 6sin²θ + i(8sinθ + 3sinθ)]/[1 + 4sin²θ]

= (4 + 6sin²θ)/(1 + 4sin²θ)  + i(11sinθ)/(1 + 4sin²θ)

Given that number is purely real,

hence imaginary part = 0

11sinθ/[1 + 4sin²θ] = 0

Since [1 + 4sin²θ] ≠0,

11sinθ = 0

sinθ = 0

or

θ = nπ, where n∈Z

Hence , given number is purely real whenever θ is equal to

integral multiple of π.

Hope, it helps

Answer 2

The above picture explains the whole concept.

I hope it will help you ☺