Find $ \left(1+a\right)^2 $ using formula.

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

where $ A = \color{blue}{ 1 } $ and $ B = \color{red}{ a }$.

$$ \begin{aligned}\left(1+a\right)^2 = \color{blue}{1^2} +2 \cdot 1 \cdot a + \color{red}{a^2} = 1+2a+a^2\end{aligned} $$

Combine like terms:

$$ \color{blue}{1} +2a+a^2+ \color{blue}{1} = a^2+2a+ \color{blue}{2} $$

Divide $ \dfrac{2+2a-2i}{a^2+2a+2} $ by $ i $ to get $ \dfrac{2a-2i+2}{a^2i+2ai+2i} $.

Step 1: To divide rational expressions, multiply the first fraction by the reciprocal of the second fraction.

Step 2: Multiply numerators and denominators.

Step 3: Simplify numerator and denominator.

$$ \begin{aligned} \frac{ \frac{2+2a-2i}{a^2+2a+2} }{i} & \xlongequal{\text{Step 1}} \frac{2+2a-2i}{a^2+2a+2} \cdot \frac{\color{blue}{1}}{\color{blue}{i}} = \\[1ex] & \xlongequal{\text{Step 2}} \frac{ \left( 2+2a-2i \right) \cdot 1 }{ \left( a^2+2a+2 \right) \cdot i } \xlongequal{\text{Step 3}} \frac{ 2+2a-2i }{ a^2i+2ai+2i } = \\[1ex] &= \frac{2a-2i+2}{a^2i+2ai+2i} \end{aligned} $$