Question

The modulus of $\frac{\left(1-i\right)}{(3+i)}+\frac{4i}{5}$ is

• A. $\sqrt{5}$ unit
• B. $\frac{\sqrt{5}}{5}$ unit
• C. $\frac{\sqrt{11}}{5}$ unit
• D. $\frac{\sqrt{12}}{5}$ unit

Answer 1

The correct option is B

$\frac{\sqrt{5}}{5}$ unit

Explanation for the correct option:

Step 1: Simplify the given complex number

In the question, a complex number $\frac{\left(1-i\right)}{(3+i)}+\frac{4i}{5}$ is given.

Simplify the given complex number as follows:

$$\begin{gathered} \frac{\left(1-i\right)}{(3+i)}+\frac{4i}{5}=\frac{\left(1-i\right)}{(3+i)}·\frac{(3-i)}{(3-i)}+\frac{4i}{5} \\ \Rightarrow \frac{\left(1-i\right)}{(3+i)}+\frac{4i}{5}=\frac{\left(1-i\right)(3-i)}{(3^2-i^2)}+\frac{4i}{5} \\ \Rightarrow \frac{\left(1-i\right)}{(3+i)}+\frac{4i}{5}=\frac{3-i-3i-1}{10}+\frac{4i}{5} \\ \Rightarrow \frac{\left(1-i\right)}{(3+i)}+\frac{4i}{5}=\frac{2-4i}{10}+\frac{4i}{5} \\ \Rightarrow \frac{\left(1-i\right)}{(3+i)}+\frac{4i}{5}=\frac{2-4i+8i}{10} \\ \Rightarrow \frac{\left(1-i\right)}{(3+i)}+\frac{4i}{5}=\frac{2+4i}{10} \\ \Rightarrow \frac{\left(1-i\right)}{(3+i)}+\frac{4i}{5}=\frac{1+2i}{5} \end{gathered}$$

So, the given complex number $\frac{\left(1-i\right)}{(3+i)}+\frac{4i}{5}$ can be written as $\frac{1+2i}{5}$.

Step 2: Find the modulus of the given complex number

Find the modulus of the given complex number as follows:

$$\begin{gathered} \left|\frac{1+2i}{5}\right|=\left|\frac{1}{5}+i\frac{2}{5}\right| \\ \Rightarrow \left|\frac{1+2i}{5}\right|=\sqrt{{\left(\frac{1}{5}\right)}^2+{\left(\frac{2}{5}\right)}^2} \\ \Rightarrow \left|\frac{1+2i}{5}\right|=\sqrt{\frac{5}{25}} \\ \Rightarrow \left|\frac{1+2i}{5}\right|=\frac{\sqrt{5}}{5} \end{gathered}$$

Therefore, the modulus of the given complex number is $\frac{\sqrt{5}}{5}$.

Hence, the correct option is $B$,