3. Given the complex number $z$ satisfies $(1-z)^{7}=-z^{7}$. Then $z=$ $\qquad$
3. $-\frac{1}{2}+\frac{1}{2} \mathrm{i} \cot \frac{k \pi}{7}(k=0,1, \cdots, 6)$.

Since $\left(1-\frac{1}{z}\right)^{7}=1$, we have,
$$
\begin{array}{l}
\frac{1}{z_{k}}=1-\cos \frac{2 k \pi}{7}-\mathrm{i} \sin \frac{2 k \pi}{7}(k=0,1, \cdots, 6) \\
\Rightarrow z_{k}=\frac{1}{1-\cos \frac{2 k \pi}{7}-\mathrm{i} \sin \frac{2 k \pi}{7}} \\
=\frac{1}{-2 \mathrm{i} \sin \frac{k \pi}{7} \cdot\left(\cos \frac{k \pi}{7}-\mathrm{i} \sin \frac{k \pi}{7}\right)} \\
=-\frac{1}{2}+\frac{1}{2} \mathrm{i} \cot \frac{k \pi}{7}(k=0,1, \cdots, 6) . \\
\end{array}
$$