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Complex Numbers — Maths STD 11 Science — Question

Maharashtra BoardEnglish MediumSTD 11 ScienceMathsComplex Numbers1 Mark
MCQ
A real value of $x$ satisfies the equation $\frac{3-4\text{ix}}{3+4\text{ix}}=\text{a}-\text{ib}\Big(\text{a},\text{b}\in\text{R}\Big),$ if $\text{a}^2+\text{b}^2=$
  • ✓
    $1$
  • B
    $-1$
  • C
    $2$
  • D
    $-2$
✓

Answer

Correct option: A.
$1$
$\text{a}-\text{ib}=\frac{3-4\text{ix}}{3+4\text{ix}}$
$=\frac{3-4\text{ix}}{3+4\text{ix}}\times\frac{3-4\text{ix}}{3-4\text{ix}}$
$=\frac{9+16\text{x}^2\text{i}^2-24\text{xi}}{9-16\text{x}^2\text{i}^2}$
$=\frac{(9-16\text{x}^2)-\text{i}(24\text{x})}{9+16\text{x}^2}$
$\Rightarrow\ |\text{a}-\text{ib}|^2=\Bigg|\frac{(9-16\text{x}^2)-\text{i}(24\text{x})}{9+16\text{x}^2}\Bigg|^2$
$\Rightarrow\text{a}^2+\text{b}^2=\frac{(9-16\text{x}^2)^2+(24\text{x})^2}{(9+16\text{x}^2)^2}$
$=\frac{81+256\text{x}^4-288\text{x}^2+576\text{x}^2}{(9+16\text{x}^2)^2}$
$=\frac{81+256\text{x}^4+288\text{x}^2}{(9+16\text{x}^2)^2}$
$=\frac{(9+16\text{x}^2)^2}{(9+16\text{x}^2)^2}$
$=1$

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