4. If the equation $x^{2}+(1-2 i) x+3 m-i=0(m$ $\in R$ ) has one real root and one complex root, then this complex root is $\qquad$

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4. $-\frac{1}{2}+2 i$.

Let the equation have a real root $\alpha$, then we have
$$
\left(\alpha^{2}+\alpha+3 m\right)+(-2 \alpha-1) i=0 .
$$

By the necessary and sufficient condition for the equality of complex numbers, we can solve to get $a=-\frac{1}{2}$. Let the other root be $\beta$, then $\alpha+\beta=2 i-1$, hence $\beta=-\frac{1}{2}+2 i$.