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BITSAT 2023Paper 2
Exam
BITSAT
Year
2023
Paper
Paper 2
Subject
Mathematics
Chapter
Complex Number

BITSAT 2023 · Paper 2 · Mathematics · Complex Number

The modulus of (1+i 3 )(2+2 i) ( 3 -i) is

The modulus of (1+i3)(2+2i)(3−i)\frac{(1+i \sqrt{3})(2+2 i)}{(\sqrt{3}-i)} is

Options

  1. A
    2
  2. B
    4
  3. C323 \sqrt{2}
  4. D222 \sqrt{2}

Correct answer

D. 222 \sqrt{2}

Step-by-step solution

(1+i3)(2+2i)3−i=2+23i+2i−233−i=(2−23)+(23+2)i3−i×3+i3+i=23−6+2i−23i+6i+23i−23−23+1=8i−84=−2+2i∴ Modulus =(−2)2+(2)2=22\begin{aligned} & \frac{(1+i \sqrt{3})(2+2 i)}{\sqrt{3}-i}=\frac{2+2 \sqrt{3} i+2 i-2 \sqrt{3}}{\sqrt{3}-i} \\ & =\frac{(2-2 \sqrt{3})+(2 \sqrt{3}+2) i}{\sqrt{3}-i} \times \frac{\sqrt{3}+i}{\sqrt{3}+i} \\ & =\frac{2 \sqrt{3}-6+2 i-2 \sqrt{3} i+6 i+2 \sqrt{3} i-2 \sqrt{3}-2}{3+1} \\ & =\frac{8 i-8}{4}=-2+2 i \\ & \therefore \text{ Modulus }=\sqrt{(-2)^2+(2)^2}=2 \sqrt{2}\end{aligned}
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