1. If the equation
$$
x^{2}+(1-2 \mathrm{i}) x+3 m-\mathrm{i}=0(m \in \mathbf{R})
$$

has one real root, then its other root is ( ).
(A) $-\frac{3}{2}+2$ i
(B) $\frac{3}{2}-2 \mathrm{i}$
(C) $-\frac{1}{2}+2$ i.
(D) $\frac{1}{2}-2 \mathrm{i}$
$-1 . C$.
Let the real root of the equation be $\alpha$, and the other root be $a+b \mathrm{i}$ $(a, b \in \mathbf{R})$. Then
$$
\begin{array}{l}
x^{2}+(1-2 \mathrm{i}) x+3 m-\mathrm{i} \\
=(x-\alpha)[x-(a+b \mathrm{i})] \\
=x^{2}-(\alpha+a+b \mathrm{i}) x+\alpha(a+b \mathrm{i})=0 .
\end{array}
$$

By comparing coefficients, we get
$$
b=2, \alpha+a=-1, \alpha b=-1 .
$$

Solving these, we find $\alpha=-\frac{1}{2}, a=-\frac{1}{2}$.