5. Let $m, n$ be non-zero real numbers, $\mathrm{i}$ be the imaginary unit, $z \in \mathbf{C}$. Then the equation
$$
|z+n \mathrm{i}|+|z-m \mathrm{i}|=n
$$

and the equation
$$
|z+n \mathrm{i}|-|z-m \mathrm{i}|=-m
$$

represent in the same complex plane (with $F_{1}, F_{2}$ as foci) ( ).

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The translation is as follows:

5. Let $m, n$ be non-zero real numbers, $\mathrm{i}$ be the imaginary unit, $z \in \mathbf{C}$. Then the equation
$$
|z+n \mathrm{i}|+|z-m \mathrm{i}|=n
$$

and the equation
$$
|z+n \mathrm{i}|-|z-m \mathrm{i}|=-m
$$

represent in the same complex plane (with $F_{1}, F_{2}$ as foci) ( ).
5. B.

Notice that, $|z+n i|+|z-m i|=n$ represents an ellipse with foci at $F_{1}(0,-n)$ and $F_{2}(0, m)$, and $n>0$; $|z+n i|-|z-m i|=-m$ represents one branch of a hyperbola with foci at $F_{1}(0,-n)$ and $F_{2}(0, m)$.
From $n=|z+n \mathrm{i}|+|z-m \mathrm{i}| \geqslant|n+m|$, we know that $m<0$.
Therefore, the branch of the hyperbola $|z+n \mathrm{i}|-|z-m \mathrm{i}|=-m$ is closer to the point $F_{2}$.