3. If $m, n$ are non-zero real numbers, and satisfy the two sets
$$
\begin{array}{l}
A=\{z|| z+n \mathrm{i}|+| z-m \mathrm{i} \mid=n, z \in \mathbf{C}\}, \\
B=\{z|| z+n \mathrm{i}|-| z-m \mathrm{i} \mid=-m, z \in \mathbf{C}\}
\end{array}
$$

are both non-empty sets, then the range of $\frac{n}{m}$ is $\qquad$ .
3. $(-\infty,-2]$.

From $n=|z+n \mathrm{i}|+|z-m \mathrm{i}|$
$$
\geqslant |(z+n \mathrm{i})-(z-m \mathrm{i})|=|m+n|,
$$

we know $n>0>m$.
$$
\begin{array}{l}
\text { Also, } -m=|z+n \mathrm{i}|-|z-m \mathrm{i}| \\
\leqslant |(z+n \mathrm{i})-(z-m \mathrm{i})|=|m+n|,
\end{array}
$$

Thus, $m+n>0$.
Hence, $-m \leqslant m+n \Rightarrow \frac{n}{m} \leqslant-2$.