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} $