Question

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

(a) 1
(b) −1
(c) 2
(d) −2

Answer 1

$$\begin{gathered} a-ib=\frac{3-4ix}{3+4ix} \\ =\frac{3-4ix}{3+4ix}\times \frac{3-4ix}{3-4ix} \\ =\frac{9+16x^2{i}^2-24xi}{9-16x^2{i}^2} \\ =\frac{\left(9-16x^2\right)-i\left(24x\right)}{9+16x^2} \\ \Rightarrow {\left|a-ib\right|}^2={\left|\frac{\left(9-16x^2\right)-i\left(24x\right)}{9+16x^2}\right|}^2 \\ \Rightarrow a^2+b^2=\frac{{\left(9-16x^2\right)}^2+{\left(24x\right)}^2}{{\left(9+16x^2\right)}^2} \\ =\frac{81+256x^4-288x^2+576x^2}{{\left(9+16x^2\right)}^2} \\ =\frac{81+256x^4+288x^2}{{\left(9+16x^2\right)}^2} \\ =\frac{{\left(9+16x^2\right)}^2}{{\left(9+16x^2\right)}^2} \\ =1 \end{gathered}$$

Hence, the correct option is (a).