Find $ \left(2+iw\right)^2 $ using formula.

$$ (A + B)^2 = \color{blue}{A^2} + 2 \cdot A \cdot B + \color{red}{B^2} $$

where $ A = \color{blue}{ 2 } $ and $ B = \color{red}{ iw }$.

$$ \begin{aligned}\left(2+iw\right)^2 = \color{blue}{2^2} +2 \cdot 2 \cdot iw + \color{red}{\left( iw \right)^2} = 4+4iw+i^2w^2\end{aligned} $$

Multiply $ \color{blue}{iw} $ by $ \left( 4+4iw+i^2w^2\right) $ $$ \color{blue}{iw} \cdot \left( 4+4iw+i^2w^2\right) = 4iw+4i^2w^2+i^3w^3 $$

Multiply $2$ by $ \dfrac{k}{4iw+4i^2w^2+i^3w^3} $ to get $ \dfrac{2k}{i^3w^3+4i^2w^2+4iw} $.

Step 1: Write $ 2 $ as a fraction by putting $ \color{red}{1} $ in the denominator.

Step 2: Multiply numerators and denominators.

Step 3: Simplify numerator and denominator.

$$ \begin{aligned} 2 \cdot \frac{k}{4iw+4i^2w^2+i^3w^3} & \xlongequal{\text{Step 1}} \frac{2}{\color{red}{1}} \cdot \frac{k}{4iw+4i^2w^2+i^3w^3} = \\[1ex] & \xlongequal{\text{Step 2}} \frac{ 2 \cdot k }{ 1 \cdot \left( 4iw+4i^2w^2+i^3w^3 \right) } \xlongequal{\text{Step 3}} \frac{ 2k }{ 4iw+4i^2w^2+i^3w^3 } = \\[1ex] &= \frac{2k}{i^3w^3+4i^2w^2+4iw} \end{aligned} $$