Question

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

• A. $\frac{4}{5}$
• B. 0
• C. $\frac{2}{5}$
• D. $-\frac{4}{5}$

Answer 1

Answer: (D)

$\frac{{\left(1+i\right)}^2}{i\left(2i-1\right)}=\frac{1+i^2+2i}{2i^2-i}$
$\Rightarrow \frac{1-1+2i}{-2-i}=-\frac{2i}{2+i}\times \frac{\left(2-i\right)}{\left(2-i\right)}=\frac{-4i+2i^2}{4-i^2}\left[\because i^2=-1\right]$
$=\frac{-4i-2}{4+1}=\frac{-2-4i}{5}\Rightarrow \frac{-2}{5}-\frac{4i}{5}$
$\therefore$ The imaginary part $=-\frac{4}{5}$