Question

$A$ real value of $x$ satisfies the equation $\left(\frac{3-4ix}{3+4ix}\right)$ $=\alpha-i\beta\left(\alpha, \beta\in R\right)$, if $\alpha^{2}+\beta^{2}$ is equal to

• A. $1$
• B. $-1$
• C. $2$
• D. $-2$

Answer 1

Answer: (A)

$\alpha-i\beta=\left(\frac{3-4ix}{3+4ix}\right)\quad\ldots\left(i\right)$
Its conjugate is given by
$\alpha+i\beta=\left(\frac{3+4ix}{3-4ix}\right)\quad\ldots\left(ii\right)$
Multiplying $\left(i\right)$ and $\left(ii\right)$, we get
$\left(\alpha-i\beta\right)\left(\alpha+i\beta\right)=\frac{3-4ix}{3+4ix}\times\frac{3+4ix}{3-4ix}$
$\Rightarrow \,\alpha^{2}+\beta^{2}=1$