Answer
$\frac{1}{2}+\frac{3i}{2}$
Work Step by Step
$\frac{2+i}{1-i}$
=$\frac{2+i}{1-i}\times\frac{1+i}{1+i}$
=$\frac{(2+i)(1+i)}{(1-i)(1+i)}$
=$\frac{2(1+i)+i(1+i)}{1^{2}-i^{2}}$
=$\frac{2+2i+i+i^{2}}{1-i^{2}}$
=$\frac{2+3i+i^{2}}{1-(-1)}$
=$\frac{2+3i-1}{2}$
=$\frac{1+3i}{2}$
=$\frac{1}{2}+\frac{3i}{2}$
