Multiply each term of $ \left( \color{blue}{z-2}\right) $ by each term in $ \left( 1-i\right) $.

$$ \left( \color{blue}{z-2}\right) \cdot \left( 1-i\right) = z-iz-2+2i $$

Subtract $i$ from $ \dfrac{z-iz-2+2i}{z+2} $ to get $ \dfrac{ \color{purple}{ -2iz+z-2 } }{ z+2 }$.

Step 1: Write $ i $ 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 $\color{blue}{z+2}$.

$$ \begin{aligned} \frac{z-iz-2+2i}{z+2} -i & \xlongequal{\text{Step 1}} \frac{z-iz-2+2i}{z+2} - \frac{i}{\color{red}{1}} = \frac{ z-iz-2+2i }{ z+2 } - \frac{ i \cdot \color{blue}{ \left( z+2 \right) }}{ 1 \cdot \color{blue}{ \left( z+2 \right) }} = \\[1ex] & \xlongequal{\text{Step 2}} \frac{ \color{purple}{ z-iz-2+2i } }{ z+2 } - \frac{ \color{purple}{ iz+2i } }{ z+2 }=\frac{ \color{purple}{ z-iz-2+2i - \left( iz+2i \right) } }{ z+2 } = \\[1ex] &=\frac{ \color{purple}{ -2iz+z-2 } }{ z+2 } \end{aligned} $$