# A real value of x satisfies the equation

Question:

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

(a) 1

(b) −1

(c) 2

(d) −2

Solution:

$a-i b=\frac{3-4 i x}{3+4 i x}$

$=\frac{3-4 i x}{3+4 i x} \times \frac{3-4 i x}{3-4 i x}$

$=\frac{9+16 x^{2} i^{2}-24 x i}{9-16 x^{2} i^{2}}$

$=\frac{\left(9-16 x^{2}\right)-i(24 x)}{9+16 x^{2}}$

$\Rightarrow|a-i b|^{2}=\left|\frac{\left(9-16 x^{2}\right)-i(24 x)}{9+16 x^{2}}\right|^{2}$

$\Rightarrow a^{2}+b^{2}=\frac{\left(9-16 x^{2}\right)^{2}+(24 x)^{2}}{\left(9+16 x^{2}\right)^{2}}$

$=\frac{81+256 x^{4}-288 x^{2}+576 x^{2}}{\left(9+16 x^{2}\right)^{2}}$

$=\frac{81+256 x^{4}+288 x^{2}}{\left(9+16 x^{2}\right)^{2}}$

$=\frac{\left(9+16 x^{2}\right)^{2}}{\left(9+16 x^{2}\right)^{2}}$

= 1

Hence, the correct option is (a).