$$
\begin{array}{l}
|z_{1}|=|z_{2}|=1, \\
z_{1}+z_{2}=\frac{1}{2}+\frac{\sqrt{3}}{2} \text { i. }
\end{array}
$$

Then $|z_{1}-z_{2}|=$ $\qquad$
2. $\sqrt{3}$.

Let $z_{1}=a+b \mathrm{i}, z_{2}=c+d \mathrm{i}$.
From the problem, we have
$$
\begin{array}{l}
a^{2}+b^{2}=1, c^{2}+d^{2}=1, \\
a+c=\frac{1}{2}, b+d=\frac{\sqrt{3}}{2} .
\end{array}
$$

Solving, we get $z_{1}=1, z_{2}=-\frac{1}{2}+\frac{\sqrt{3}}{2} \mathrm{i}$ or
$$
z_{2}=1, z_{1}=-\frac{1}{2}+\frac{\sqrt{3}}{2} \mathrm{i}.
$$

Therefore, $\left|z_{1}-z_{2}\right|=\sqrt{3}$.