3. Given the complex number sequence $\left\{z_{n}\right\}$ satisfies $z_{1}=1, z_{n+1}=\overline{z_{n}}+1+n i(n=1,2, \cdots)$, then $z_{2015}=$
Answer: $2015+1007 \mathrm{i}$. Solution: From the given, for all positive integers $\mathrm{n}$, we have
$$
z_{n+2}=\overline{z_{n+1}}+1+(n+1) i=\overline{z_{n}+1+n i}+1+(n+1) i=z_{n}+2+i \text {, }
$$

Thus, $z_{2015}=z_{1}+1007 \times(2+i)=2015+1007 i$.