Add $ \dfrac{2+i}{z-i} $ and $ \dfrac{2-i}{z+i} $ to get $ \dfrac{ \color{purple}{ 2i^2+4z } }{ -i^2+z^2 }$.

To add raitonal expressions, both fractions must have the same denominator.
We can create a common denominator by multiplying the first fraction by $ \color{blue}{ i+z }$ and the second by $\color{blue}{ -i+z }$.

$$ \begin{aligned} \frac{2+i}{z-i} + \frac{2-i}{z+i} & = \frac{ \left( 2+i \right) \cdot \color{blue}{ \left( i+z \right) }}{ \left( z-i \right) \cdot \color{blue}{ \left( i+z \right) }} + \frac{ \left( 2-i \right) \cdot \color{blue}{ \left( -i+z \right) }}{ \left( z+i \right) \cdot \color{blue}{ \left( -i+z \right) }} = \\[1ex] &=\frac{ \color{purple}{ 2i+2z+i^2+iz } }{ \cancel{iz}+z^2-i^2 -\cancel{iz} } + \frac{ \color{purple}{ -2i+2z+i^2-iz } }{ \cancel{iz}+z^2-i^2 -\cancel{iz} } = \\[1ex] &=\frac{ \color{purple}{ 2i^2+4z } }{ -i^2+z^2 } \end{aligned} $$

Multiply $ \dfrac{1}{2} $ by $ \dfrac{2i^2+4z}{-i^2+z^2} $ to get $ \dfrac{ 2i^2+4z }{ -2i^2+2z^2 } $.

Step 1: Multiply numerators and denominators.

Step 2: Simplify numerator and denominator.

$$ \begin{aligned} \frac{1}{2} \cdot \frac{2i^2+4z}{-i^2+z^2} & \xlongequal{\text{Step 1}} \frac{ 1 \cdot \left( 2i^2+4z \right) }{ 2 \cdot \left( -i^2+z^2 \right) } = \\[1ex] & \xlongequal{\text{Step 2}} \frac{ 2i^2+4z }{ -2i^2+2z^2 } \end{aligned} $$