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 $ \pm1 $ 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 $ \pm1\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} $ .