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

The correct option is D

$\frac{-4}{5}$

Explanation for the correct option:

Step 1: Simplify the given complex expression

We can simplify $\frac{{\left(1+i\right)}^2}{i\left(2i-1\right)}$ as

$\frac{{\left(1+i\right)}^2}{i\left(2i-1\right)}=\frac{1+2i+i^2}{2i^2-i}$

$=\frac{1+2i-1}{-2-i}$ $\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{[}\mathbf{\because }{\mathbf{i}}^{\mathbf{2}}\mathbf{=}\mathbf{-}\mathbf{1}\mathbf{]}$

$\therefore$$\frac{{\left(1+i\right)}^2}{i\left(2i-1\right)}$ $=\frac{-2i}{2+i}$

Step 2: Multiply and divide by $(2-i)$

$\frac{{\left(1+i\right)}^2}{i\left(2i-1\right)}$$=\frac{-2i}{2+i}\times \frac{2-i}{2-i}$

$=\frac{-4i+2i^2}{4-i^2}$

$=\frac{-4i+2(-1)}{4-(-1)}$ $\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{[}\mathbf{\because }{\mathbf{i}}^{\mathbf{2}}\mathbf{=}\mathbf{-}\mathbf{1}\mathbf{]}$

$=\frac{-2-4i}{5}$

$\therefore$ $\frac{{\left(1+i\right)}^2}{i\left(2i-1\right)}$$=\frac{-2}{5}-\frac{4}{5}i$

Step 3: Compare the simplified form of given complex expression with $a+ib$

On comparing $\frac{-2}{5}-\frac{4}{5}i$ with $a+ib$ we get,

$a=\frac{-2}{5}$ and $b=\frac{-4}{5}$, where $a$ is the real part and $b$is the imaginary part

So, the imaginary part of the given complex number is $\frac{-4}{5}$

Hence, option (D) is the correct answer.