Answer
$1-2i$
Work Step by Step
Multiply both the denominator and the numerator by $-i$, which is the conjugate of $i$, then use the fact that $i^2=-1$ to obtain:
\begin{align*}
&=\frac{2+i}{i}\cdot\frac{-i}{-i}\\
\\&=\frac{(2+i)(-i)}{-i^2}\\
\\&=\frac{-2i-i^2}{-(-1)}\\
\\&=\frac{-2i-(-1)}{1}\\
\\&=\frac{-2i+1}{1}\\
\\&=1-2i
\end{align*}
