Question

The real part of $\frac{{\left(1+i\right)}^2}{\left(3-i\right)}$ is

• A. $\frac{1}{3}$
• B. $\frac{1}{5}$
• C. $-\frac{1}{3}$
• D. None of these

Answer 1

Answer: (D)

${\left(1+i\right)}^2=1+i^2+2i=2i$
$\therefore \frac{{\left(1+i\right)}^2}{3-i}$
$=\frac{2i\left(3+i\right)}{3^2-i^2}$
$=\frac{6i-2}{10}$
$=\frac{-1+3i}{5}$
$\therefore$ Real part $=\frac{-1}{5}$