Q74 / 180 · KCET 2023 · Mathematics
The modulus of the complex number
$\frac{(1+i)^2(1+3 i)}{(2-6 i)(2-2 i)}$ is
- A$\frac{2}{\sqrt{2}}$
- B$\frac{1}{\sqrt{2}}$
- Correct answerC$\frac{\sqrt{2}}{4}$
- D$\frac{4}{\sqrt{2}}$
Solution
Here,
$\frac{(1+i)^2(1+3 i)}{(2-6 i)(2-2 i)}$
$\begin{aligned} & =\frac{\left(1+i^2+2 i\right)(1+3 i)}{4-4 i-12 i+12 i^2} \\ & =\frac{2 i+6 i^2}{-8-16 i}=\frac{2 i-6}{-8-16 i} \\ & =\frac{i-3}{-4-8 i} \\ & =\frac{3-i}{(4+8 i)} \times \frac{(4-8 i)}{(4-8 i)} \\ & =\frac{12-28 i-8}{16+64} \\ & =\frac{4-28 i}{80}=\frac{1-7 i}{20} \\ & =\frac{1}{20}-\frac{7 i}{20}\end{aligned}$
Let $z=x+i y$
$\because|z|=\sqrt{x^2+y^2}$
$\begin{aligned} & \Rightarrow|z|=\sqrt{\left(\frac{1}{20}\right)^2+\left(\frac{7}{20}\right)^2} \\ & \Rightarrow|z|=\frac{1}{20} \sqrt{50}=\frac{5}{20} \sqrt{2} \\ & \Rightarrow|z|=\frac{\sqrt{2}}{4}\end{aligned}$
Hence, the correct answer is option: 'C'.