A real value of x satisfies the equation 3-4 i x-3-4 i x-a-ba- b - R- if a2-b2

The correct option is A 1 Given equation, (3 4ix/3+4ix )=a b(α, βϵ R) ⇒ [3 4ix/3+4ix ]=α iβ Now (x iβ) =(3 4ix)(3 4ix)/(3+4ix)(3 4ix)=9+16i^2x^2 24ix/9 16i^2x ...

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Standard XII

Mathematics

Complex Numbers

A real value ...

Question

A real value of x satisfies the equation $\frac{3 - 4 i x}{3 + 4 i x} = a - b (a, b ϵ R), if a^{2} + b^{2}$

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Solution

The correct option is A
1

Given equation,

$(\frac{3 - 4 i x}{3 + 4 i x}) = a - b (α, β ϵ R) \Rightarrow [\frac{3 - 4 i x}{3 + 4 i x}] = α - i β Now (x - i β) = \frac{(3 - 4 i x) (3 - 4 i x)}{(3 + 4 i x) (3 - 4 i x)} = \frac{9 + 16 i^{2} x^{2} - 24 i x}{9 - 16 i^{2} x^{2}} \Rightarrow α - i β = \frac{9 - 16 x^{2} - 24 i x}{9 + 16 x^{2}} \Rightarrow α - i β = \frac{9 - 16 x^{2}}{9 + 16 x^{2}} - \frac{i 24 x}{9 + 16 x^{2}} . . . (i) ∴ α + i β = \frac{9 - 16 x^{2}}{9 + 16 x^{2}} + \frac{i 24 x}{9 + 16 x^{2}} . . . (i i) So, (α - i β) (α + i β) = {(\frac{9 - 16 x^{2}}{9 + 16 x^{2}})}^{2} - {(\frac{i 24 x}{9 + 16 x^{2}})}^{2} ∴ α^{2} + β^{2} = \frac{81 + 256 x^{4} - 288 x^{2} + 576 x^{2}}{(9 + 16 x^{2})^{2}} = \frac{81 + 256 x^{4} + 288 x^{2}}{(9 + 16 x^{2})^{2}} = \frac{(9 + 16 x^{2})^{2}}{(9 + 16 x^{2})^{2}} = 1$

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Similar questions

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A real value of x satisfies the equation 3−4ix3+4ix=a−b(a, bϵR), if a2+b2

A real value of x satisfies the equation $\frac{3 - 4 i x}{3 + 4 i x} = a - b (a, b ϵ R), if a^{2} + b^{2}$