Find $ \left(0+2j\right)^2 $ using formula.

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

where $ A = \color{blue}{ 0 } $ and $ B = \color{red}{ 2j }$.

$$ \begin{aligned}\left(0+2j\right)^2 = \color{blue}{0^2} +2 \cdot 0 \cdot 2j + \color{red}{\left( 2j \right)^2} = 00j+4j^2\end{aligned} $$

Subtract $ \dfrac{1}{0+2j} $ from $ \dfrac{-1}{00j+4j^2} $ to get $ \dfrac{ \color{purple}{ -4j^2-2j } }{ 8j^3 }$.

To subtract raitonal expressions, both fractions must have the same denominator.
We can create a common denominator by multiplying the first fraction by $ \color{blue}{ 2j }$ and the second by $\color{blue}{ 4j^2 }$.

$$ \begin{aligned} \frac{-1}{00j+4j^2} - \frac{1}{0+2j} & = \frac{ \left( -1 \right) \cdot \color{blue}{ 2j }}{ \left( 00j+4j^2 \right) \cdot \color{blue}{ 2j }} - \frac{ 1 \cdot \color{blue}{ 4j^2 }}{ \left( 0+2j \right) \cdot \color{blue}{ 4j^2 }} = \\[1ex] &=\frac{ \color{purple}{ -2j } }{ 0j0j^2+8j^3 } - \frac{ \color{purple}{ 4j^2 } }{ 0j0j^2+8j^3 }=\frac{ \color{purple}{ -4j^2-2j } }{ 8j^3 } \end{aligned} $$

Subtract $1$ from $ \dfrac{-4j^2-2j}{8j^3} $ to get $ \dfrac{ \color{purple}{ -8j^3-4j^2-2j } }{ 8j^3 }$.

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

Step 2: To subtract raitonal expressions, both fractions must have the same denominator.
We can create a common denominator by multiplying the second fraction by $\color{blue}{8j^3}$.

$$ \begin{aligned} \frac{-4j^2-2j}{8j^3} -1 & \xlongequal{\text{Step 1}} \frac{-4j^2-2j}{8j^3} - \frac{1}{\color{red}{1}} = \frac{ -4j^2-2j }{ 8j^3 } - \frac{ 1 \cdot \color{blue}{ 8j^3 }}{ 1 \cdot \color{blue}{ 8j^3 }} = \\[1ex] & \xlongequal{\text{Step 2}} \frac{ \color{purple}{ -4j^2-2j } }{ 8j^3 } - \frac{ \color{purple}{ 8j^3 } }{ 8j^3 }=\frac{ \color{purple}{ -8j^3-4j^2-2j } }{ 8j^3 } \end{aligned} $$