Question

A real value of x satisfies the equation `((3 - 4ix)/(3 + 4ix))` = α − iβ (α, β ∈ R) if α² + β² = ______.

विकल्प

• 1

• –1

• 2

• –2

Answer 1

A real value of x satisfies the equation `((3 - 4ix)/(3 + 4ix))` = α − iβ (α, β ∈ R) if α² + β² = 1.

Explanation:

Given that: `((3 - 4ix)/(3 + 4ix))` = α − iβ

⇒ `((3 - 4ix)/(3 + 4ix) xx (3 - 4ix)/(3 - 4ix))` = α − iβ

⇒ `(9 - 12ix - 12ix + 16i^2 x^2)/(9 - 16i^2 x^2)` = α − iβ 

⇒ `(9 - 24ix - 16x^2)/(9 + 16x^2)` = α − iβ 

⇒ `(9 - 16x^2)/(9 + 16x^2) - (24x)/(9 + 16x^2) i` = α − iβ   .....(i)

⇒ `(9 - 16x^2)/(9 + 16x^2) + (24x)/(9 + 16x^2) i` = α + iβ

Multiplying equation (i) and (ii) we get

⇒ `((9 - 16x^2)/(9 + 16x^2))^2 + ((24x)/(9 + 16x^2))^2` = α² + β²

⇒ `((9 - 16x^2)^2 + (24x)^2)/(9 + 16x^2)^2` = α² + β²

⇒ `(81 + 256x^4 - 288x^2 + 576x^2)/(9 + 16x^2)^2` = α² + β²

⇒ `(81 + 256x^4 + 288x^2)/(9 + 16x^2)^2` = α² + β²

⇒ `(9 + 16x^2)^2/(9 + 16x^2)^2` = α² + β²

So, = α² + β² = 1