Question

$\text{Re}\frac{{\left(1-i\right)}^2}{\left(3-i\right)}$ is equal to

• A. $-\frac{1}{5}$
• B. $\frac{1}{5}$
• C. $\frac{1}{10}$
• D. $-\frac{1}{10}$

Answer 1

The correct option is B

$\frac{1}{5}$

Explanation for the correct option:

Given expression is $\text{Re}\frac{{\left(1-i\right)}^2}{\left(3-i\right)}$ let us simplify it and finding real part of it

$$\begin{gathered} \frac{{\left(1-i\right)}^2}{3-i}=\frac{\left(1-2i-1\right)}{\left(3-i\right)} \\ =\frac{-2i}{\left(3-i\right)} \end{gathered}$$ On rationalising we get,

$$\begin{gathered} =\frac{-2i(3+i)}{(3+i).(3-i)}(i^2=-1),(a+b).(a-b)=a^2-b^2 \\ =\frac{(-6i+2)}{(9+1)} \\ =\frac{(2-6i)}{10} \\ =\frac{1}{5}-\frac{3}{5}i \end{gathered}$$

So, we see that the real part of given expression is $\frac{1}{5}$

Hence (B) is the correct option