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Complex Numbers — Maths STD 11 Science — Question

CBSE BoardEnglish MediumSTD 11 ScienceMathsComplex Numbers1 Mark
MCQ
If $(1+\text{i})(1+2\text{i})(1+3\text{i})...(1+\text{ni})=\text{a}+\text{ib},$ then $2\times5\times10\times...\times(1+\text{n}^2)$ is equal to:
  • A
    $\sqrt{\text{a}^2+\text{b}^2}$
  • B
    $\sqrt{\text{a}^2-\text{b}^2}$
  • ✓
    $\text{a}^2+\text{b}^2$
  • D
    $\text{a}^2-\text{b}^2$
✓

Answer

Correct option: C.
$\text{a}^2+\text{b}^2$
$(1+\text{i})(1+2\text{i})(1+3\text{i})...(1+\text{ni})=\text{a}+\text{ib}$
Taking modulus on both the sides, we get:
$|(1+\text{i})(1+2\text{i})(1+3\text{i})...(1+\text{ni})|=|\text{a}+\text{ib}|$
$|(1+\text{i})(1+2\text{i})(1+3\text{i})...(1+\text{ni})|$ can be written as $|(1+\text{i})||(1+2\text{i})||(1+3\text{i})|...|(1+\text{ni})|$
$\sqrt{1^2+1^2}\times\sqrt{1^2+2^2}\times\sqrt{1^2+3^2}\times...\times\sqrt{1+\text{n}^2}=\sqrt{\text{a}^2+\text{b}^2}$
$\Rightarrow\sqrt{2}\times\sqrt{5}\times\sqrt{10}\times...\times\sqrt{1+\text{n}^2}=\sqrt{\text{a}^2+\text{b}^2}$
Squaring on both the sides, we get:
$2\times5\times10\times...\times(1+\text{n}^2)=\text{a}^2+\text{b}^2$

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