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

Answer

  1. 1

Solution:

$\frac{1-\text{ix}}{1+\text{ix}}=\text{a}+\text{ib}$

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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