3. (Mathematics for Middle School, 2004, Issue 4, Olympiad Training Problem) Let $z=m+n \mathrm{i}$ be a complex root of the equation $a z^{4}+\mathrm{i} b z^{2}+$ $c z^{2}+\mathrm{i} d z+e=0$, 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. $m-n \mathrm{i}$
B. $-m+n \mathrm{i}$
C. $n+m \mathrm{i}$
D. $1-m+n \mathrm{i}$
3. B. Reason: Since $z=m+n \mathrm{i}$ is a root of the known equation, then, $a(\mathrm{i} z)^{4}-b(\mathrm{i} z)^{3}-c(\mathrm{i} z)^{2}+d(\mathrm{i} z)+e=0$

i.e., $z^{\prime}=(m+n \mathrm{i}) \cdot \mathrm{i}=-n+m \mathrm{i}$ 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 real-coefficient equation appear in pairs, $\vec{z}=-n-m \mathrm{i}$ is also a root of (*).
Accordingly, $\frac{1}{\mathrm{i}}(-n-m \mathrm{i})=-m+n \mathrm{i}$ is a root of the original equation.