Answer
$\frac{12-i}{13}$
Work Step by Step
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}$
