In the ring of $\mathbb Z[i]$ divide $7+2i$ by $2+i$. In the ring of $\mathbb Z[i]$ use our division algorithm divide $7+2i$ by $2+i$.
$$\frac{7+2i}{2+i}\cdot \frac{2-i}{2-i}=\frac{16-3i}{5}=\frac{16}{5}-\frac{3}{5}i$$
Does this mean you cannot divide $7+2i$ by $2+i$, because the real and imaginary parts are $\frac{16}{5}$ and $\frac{3}{5}i$ which are not integers.

Presumably, what's meant is "divide with remainder". So you have the quotient $(16/5)-(3/5)i$, which, rounding everything to the nearest integer, is $3-i$, with remainder $(1/5)+(2/5)i$. That is, $${7+2i\over2+i}=(3-i)+({1\over5}+{2\over5}i)$$ Multiply through by $2+i$ to get the nicer-looking $$7+2i=(3-i)(2+i)+i$$
