## Algebra and Trigonometry 10th Edition

$\frac{12-i}{13}$
Using the rules of imaginary numbers: $\frac{4}{2-3i} = \frac{4}{2-3i} * \frac{2+3i}{2+3i} = \frac{8+12i}{4-9i^2}=\frac{8+12i}{13}$ $\frac{2}{1+i}=\frac{2}{1+i}*\frac{1-i}{1-i}=\frac{2-2i}{1-i^2}=1-i$ Hence, the final expression is: $\frac{8+12i}{13}+1-i = \frac{8+12i+13-13i}{13}=$$\frac{12-i}{13}$