10. If the complex numbers $z_{1}$ and $z_{2}$ satisfy $\left|z_{1}\right|=\left|z_{2}\right|$, and $z_{1}-z_{2}=2-\mathrm{i}$, then the value of $\frac{z_{1} z_{2}}{\left|z_{1} z_{2}\right|}$ is $\qquad$.
10. $-\frac{3}{5}+\frac{4}{5}$ i.

Notice that $\left|z_{1} z_{2}\right|=\left|z_{1}\right|^{2}=\left|z_{2}\right|^{2}$, then
$$
\begin{array}{l}
2-\mathrm{i}=z_{1}-z_{2}=\frac{1}{\left|z_{2} \overline{z_{2}}\right|} z_{1} z_{2} \overline{z_{2}}-\frac{1}{\left|z_{1} \overline{z_{1}}\right|} z_{2} z_{1} \overline{z_{1}} \\
=\frac{z_{1} z_{2}}{\left|z_{1} z_{2}\right|}\left(\overline{z_{2}}-\overline{z_{1}}\right)=\frac{z_{1} z_{2}}{\left|z_{1} z_{2}\right|}\left(\overline{z_{2}-z_{1}}\right) \\
=-\frac{z_{1} z_{2}}{\left|z_{1} z_{2}\right|}(2+\mathrm{i}) .
\end{array}
$$

Thus, $\frac{z_{1} z_{2}}{\left|z_{1} z_{2}\right|}=-\frac{2-\mathrm{i}}{2+\mathrm{i}}=-\frac{3}{5}+\frac{4}{5} \mathrm{i}$.