14. Let the complex numbers $z_{1}, z_{2}$ satisfy
$$
\begin{array}{l}
\left|z_{1}\right|=\left|z_{2}\right|=1 \text {, and } z_{1}+z_{2}-1=0 \text {. } \\
\text { Then }\left(z_{1}-z_{2}\right)^{2}=
\end{array}
$$
14. -3 .

From the problem, we know that the points corresponding to $z_{1}$, $z_{2}$, and $-1$ are the three equally divided points on the unit circle. Without loss of generality, let's assume the point corresponding to $z_{1}$ is in the first quadrant. Then,
$$
\begin{array}{l}
z_{1}=\cos \frac{\pi}{3}+\mathrm{i} \sin \frac{\pi}{3}=\frac{1}{2}+\frac{\sqrt{3}}{2} \mathrm{i}, \\
z_{2}=\overline{z_{1}}=\frac{1}{2}-\frac{\sqrt{3}}{2} \mathrm{i} .
\end{array}
$$

Thus, $\left(z_{1}-z_{2}\right)^{2}=(\sqrt{3} \mathrm{i})^{2}=-3$.