2. Let $z=m+n \mathrm{i}$ be a complex root of the equation
$$
a z^{4}+\mathrm{i} b z^{3}+c z^{2}+\mathrm{i} d z+e=0
$$

where $a, b, c, d, e, m, n \in \mathbf{R}$. Then which of the following numbers must also be a root of the equation? ( ).
(A) $m-n$ i
(B) $-m+n$ i
(C) $n+m$ i
(D) $1-m+n \mathrm{i}$
2.B.

Since $z=m+n$ i is a root of the known equation, we have
$$
a(\mathrm{i} z)^{4}-b(\mathrm{i} z)^{3}-c(\mathrm{i} z)^{2}+d(\mathrm{i} z)+e=0 \text {. }
$$

This means that $z^{\prime}:(m+n \mathrm{i}) \mathrm{i}=-n+m \mathrm{i}$ is a root of the equation
$$
a x^{4}-b x^{3}-c x^{2}+d x+e=0
$$

Since the complex roots of a polynomial equation with real coefficients occur in conjugate pairs, $\overline{z^{\prime}}=-n$ $-m$ i is also a root of (1).

Accordingly, $\frac{1}{\mathrm{i}}(-n-m \mathrm{i})=-m+n \mathrm{i}$ is a root of the original equation.