Divide $ \, -4-i \, $ by $ \, 1+i \, $ to get $\,\, \dfrac{-5+3i}{2} $.
( view steps )

Find $ \left(-1-i\right)^2 $ in two steps.

S1: Change all signs inside bracket.

S2: Apply 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}{ i }$.

$$ \begin{aligned}\left(-1-i\right)^2& \xlongequal{ S1 } \left(1+i\right)^2 \xlongequal{ S2 } \color{blue}{1^2} +2 \cdot 1 \cdot i + \color{red}{i^2} = \\[1 em] & = 1+2i+i^2\end{aligned} $$

Multiply $-4+i$ by $ \dfrac{-5+3i}{2} $ to get $ \dfrac{3i^2-17i+20}{2} $.

Step 1: Write $ -4+i $ as a fraction by putting $ \color{red}{1} $ in the denominator.

Step 2: Multiply numerators and denominators.

Step 3: Simplify numerator and denominator.

$$ \begin{aligned} -4+i \cdot \frac{-5+3i}{2} & \xlongequal{\text{Step 1}} \frac{-4+i}{\color{red}{1}} \cdot \frac{-5+3i}{2} \xlongequal{\text{Step 2}} \frac{ \left( -4+i \right) \cdot \left( -5+3i \right) }{ 1 \cdot 2 } = \\[1ex] & \xlongequal{\text{Step 3}} \frac{ 20-12i-5i+3i^2 }{ 2 } = \frac{3i^2-17i+20}{2} \end{aligned} $$

$$ i^2 = -1 $$

$$ 3i^2 = 3 \cdot (-1) = -3 $$

Combine like terms:

$$ \, \color{blue}{ \cancel{1}} \,+2i \, \color{blue}{ -\cancel{1}} \, = 2i $$

Combine like terms:

$$ \color{blue}{-3} -17i+ \color{blue}{20} = -17i+ \color{blue}{17} $$

Divide $ \, 1+3i \, $ by $ \, 2i \, $ to get $\,\, \dfrac{3-i}{2} $.
( view steps )

Add $ \dfrac{3-i}{2} $ and $ \dfrac{-17i+17}{2} $ to get $ \dfrac{-18i+20}{2} $.

To add expressions with the same denominators, we add the numerators and write the result over the common denominator.

$$ \begin{aligned} \frac{3-i}{2} + \frac{-17i+17}{2} & = \frac{3-i}{\color{blue}{2}} + \frac{-17i+17}{\color{blue}{2}} =\frac{ 3-i + \left( -17i+17 \right) }{ \color{blue}{ 2 }} = \\[1ex] &= \frac{-18i+20}{2} \end{aligned} $$