College Algebra (10th Edition)

$1-2i$
Recall, $i=\sqrt{-1}$, so $i^{2}=-1$. Thus, we multiply by $\frac{-i}{-i}$ to obtain: $\displaystyle \frac{2+i}{i}$ $\displaystyle =\frac{2+i}{i}*\frac{-i}{-i}$ $\displaystyle=\frac{-i(2+i)}{-i*i}$ $\displaystyle =\frac{-2i-i^{2}}{-i^{2}}$ $\displaystyle=\frac{-2i-(-1)}{-(-1)}$ $\displaystyle=\frac{-2i+1}{1}=1-2i$