Subtract $ \dfrac{2}{1+i} $ from $ \dfrac{1}{1-4i} $ to get $ \dfrac{ \color{purple}{ 9i-1 } }{ -4i^2-3i+1 }$.

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}{ i+1 }$ and the second by $\color{blue}{ -4i+1 }$.

$$ \begin{aligned} \frac{1}{1-4i} - \frac{2}{1+i} & = \frac{ 1 \cdot \color{blue}{ \left( i+1 \right) }}{ \left( 1-4i \right) \cdot \color{blue}{ \left( i+1 \right) }} - \frac{ 2 \cdot \color{blue}{ \left( -4i+1 \right) }}{ \left( 1+i \right) \cdot \color{blue}{ \left( -4i+1 \right) }} = \\[1ex] &=\frac{ \color{purple}{ i+1 } }{ i+1-4i^2-4i } - \frac{ \color{purple}{ -8i+2 } }{ i+1-4i^2-4i }=\frac{ \color{purple}{ i+1 - \left( -8i+2 \right) } }{ -4i^2-3i+1 } = \\[1ex] &=\frac{ \color{purple}{ 9i-1 } }{ -4i^2-3i+1 } \end{aligned} $$

$$ -4i^2 = -4 \cdot (-1) = 4 $$

Simplify denominator

$$ \color{blue}{4} -3i+ \color{blue}{1} = -3i+ \color{blue}{5} $$

Divide $ \, -1+9i \, $ by $ \, 5-3i \, $ to get $\,\, \dfrac{-16+21i}{17} $.
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