• # question_answer The imaginary part of  $\frac{{{(1+i)}^{2}}}{(2-i)}$is A) $\frac{1}{5}$ B) $\frac{3}{5}$ C) $\frac{4}{5}$ D) None of these

We have$\frac{{{(1+i)}^{2}}}{2-i}=\frac{(2i)(2+i)}{(2-i)(2+i)}$$=-\frac{2}{5}+i\frac{4}{5}$. Thus Im $(z)=\frac{4}{5}$.