jmc

Because $u$ and $v$ have integer parts, $|u{|}^2$ and $|v{|}^2$ are nonnegative integers. From $uv=10$, it follows that $|u{|}^2\cdot |v{|}^2=100$. So $|u{|}^2$ and $|v{|}^2$ are positive integers whose product is $100$. We will divide the count into three cases: $|u|<|v|$, $|u|=|v|$, and $|u|>|v|$.

Let’s handle the case $|u|<|v|$ first. In that case, $|u{|}^2$ is a small divisor of $100$: either $1,2,4$, or $5$.

If $|u{|}^2=1$, then we have $4$ choices for $u$: either $\pm 1$ or $\pm i$.

If $|u{|}^2=2$, then we have $4$ choices: $\pm 1\pm i$.

If $|u{|}^2=4$, then we have $4$ choices: $\pm 2$ or $\pm 2i$.

If $|u{|}^2=5$, then we have $8$ choices: $\pm 1\pm 2i$ or $\pm 2\pm i$.

Altogether, we have $20$ choices for $u$. Each such choice gives a single valid choice for $v$, namely $v=\frac{10}{u}=\frac{10\overline{u}}{|u{|}^2}$. So we have $20$ pairs in the case $|u|<|v|$.

Let’s next handle the case $|u|=|v|$. In that case, $|u{|}^2=|v{|}^2=10$. So we have $8$ choices for $u$: either $\pm 1\pm 3i$ or $\pm 3\pm i$. Each such choice determines $v$, namely $v=10/u=u$. So we have $8$ pairs in the case $|u|=|v|$.

Finally, we have the case $|u|>|v|$. By symmetry, it has the same count as the first case $|u|<|v|$. So we have $20$ pairs in this case.

Altogether, the number of pairs is $20+8+20$, which is $\boxed{48}$ .