## Algebra and Trigonometry 10th Edition

$\frac{9+83i}{85}$
Using the rules of imaginary numbers: $\frac{1}{2+i} = \frac{2-i}{2-i}=\frac{2-i}{2^2-i^2}=\frac{2-i}{5}$ $\frac{5}{1+4i} = \frac{5}{1+4i}*\frac{1-4i}{1-4i}=\frac{5-20i}{1-16i^2}=\frac{5-20i}{17}$ Write the final expression: $\frac{2-i}{5}-\frac{5-20i}{17} = \frac{17(2-i)+5(5-20i)}{17*5} =$$\frac{9+83i}{85}$