A real value of x satisfies the equation ((3-4ix/3+4ix)) =α-iβ(α, β∈ R), if α2+β2 is equal to

Question

A real value of x satisfies the equation ((3-4ix/3+4ix)) =α-iβ(α, β∈ R), if α2+β2 is equal to (A) 1 (B) -1 (C) 2 (D) -2

Tardigrade Admin · Accepted Answer

α-iβ=((3-4ix/3+4ix)) ldots(i) Its conjugate is given by α+iβ=((3+4ix/3-4ix)) ldots(ii) Multiplying (i) and (ii), we get (α-iβ)(α+iβ)=(3-4ix/3+4ix)×(3+4ix/3-4ix) ⇒ α2+β2=1

Q. $A$ real value of $x$ satisfies the equation $(\frac{3 - 4 i x}{3 + 4 i x})$ $= α - i β (α, β \in R)$ , if $α^{2} + β^{2}$ is equal to

$1$

$- 1$

$2$

$- 2$

$α - i β = (\frac{3 - 4 i x}{3 + 4 i x}) \dots (i)$
Its conjugate is given by
$α + i β = (\frac{3 + 4 i x}{3 - 4 i x}) \dots (ii)$
Multiplying $(i)$ and $(ii)$ , we get
$(α - i β) (α + i β) = \frac{3 - 4 i x}{3 + 4 i x} \times \frac{3 + 4 i x}{3 - 4 i x}$
$\Rightarrow α^{2} + β^{2} = 1$

Q. A real value of x satisfies the equation (3+4ix3−4ix​) =α−iβ(α,β∈R), if α2+β2 is equal to

1

−1

2

−2

α−iβ=(3+4ix3−4ix​)…(i) Its conjugate is given by α+iβ=(3−4ix3+4ix​)…(ii) Multiplying (i) and (ii), we get (α−iβ)(α+iβ)=3+4ix3−4ix​×3−4ix3+4ix​ ⇒α2+β2=1

Q. $A$ real value of $x$ satisfies the equation $(\frac{3 - 4 i x}{3 + 4 i x})$ $= α - i β (α, β \in R)$ , if $α^{2} + β^{2}$ is equal to

$1$

$- 1$

$2$

$- 2$

$α - i β = (\frac{3 - 4 i x}{3 + 4 i x}) \dots (i)$
Its conjugate is given by
$α + i β = (\frac{3 + 4 i x}{3 - 4 i x}) \dots (ii)$
Multiplying $(i)$ and $(ii)$ , we get
$(α - i β) (α + i β) = \frac{3 - 4 i x}{3 + 4 i x} \times \frac{3 + 4 i x}{3 - 4 i x}$
$\Rightarrow α^{2} + β^{2} = 1$