3. Given the complex number $z$ satisfies $(1-z)^{7}=-z^{7}$. Then $z=$ $\qquad$ .
$\begin{array}{l}\text { 3. }-\frac{1}{2}+\frac{1}{2} \mathrm{i} \cot \frac{k \pi}{7}(k=0,1, \cdots, 6) . \\ \text { Since }\left(1-\frac{1}{z}\right)^{7}=1, \text { therefore, } \\ \frac{1}{z_{k}}=1-\cos \frac{2 k \pi}{7}-\mathrm{i} \sin \frac{2 k \pi}{7}(k=0,1, \cdots, 6) \\ \Rightarrow z_{k}=\frac{1}{1-\cos \frac{2 k \pi}{7}-\mathrm{i} \sin \frac{2 k \pi}{7}} \\ =\frac{1}{-2 \mathrm{i} \sin \frac{k \pi}{7} \cdot\left(\cos \frac{k \pi}{7}-\mathrm{i} \sin \frac{k \pi}{7}\right)} \\ =-\frac{1}{2}+\frac{1}{2} \mathrm{i} \cot \frac{k \pi}{7}(k=0,1, \cdots, 6) .\end{array}$