MCQ
If $\frac{1-\text{ix}}{1+\text{ix}}=\text{a}+\text{ib},$ then $\text{a}^2+\text{b}^2=$
  • $1$
  • B
    $-1$
  • C
    $0$
  • D
    none of these

Answer

Correct option: A.
$1$
$\frac{1-\text{ix}}{1+\text{ix}}=\text{a}+\text{ib}$
Taking modulus on both the sides, we get:
$\Big|\frac{1-\text{ix}}{1+\text{ix}}\Big|=\big|\text{a}+\text{ib}\big|$
$\Rightarrow\frac{\sqrt{1^2+\text{x}^2}}{\sqrt{1^2+\text{x}^2}}=\sqrt{\text{a}^2+\text{b}^2}$
$\Rightarrow\sqrt{\text{a}^2+\text{b}^2}=1$
Squaring both the sides, we get:
$\text{a}^2+\text{b}^2=1$

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