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Question
A real value of x satisfies the equation `((3 - 4ix)/(3 + 4ix))` = α − iβ (α, β ∈ R) if α2 + β2 = ______.
Options
1
–1
2
–2
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Solution
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
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