3. Given the complex number sequence $\left\{z_{n}\right\}$ satisfies $z_{1}=1, z_{n+1}=\overline{z_{n}}+1+n \mathrm{i}(n=1,2,3 \cdots)$, where $\mathrm{i}$ is the imaginary unit, $\overline{z_{n}}$ represents the conjugate of $z_{n}$, then the value of $z_{2015}$ is $\qquad$.
$$
\begin{array}{l}
\text { Let } z_{n}=a_{n}+b_{n} \mathrm{i}\left(a_{n}, b_{n} \in \mathbf{R}\right), \text { then } a_{n+1}+b_{n+1} \mathrm{i}=a_{n}-b_{n} \mathrm{i}+1+n \mathrm{i} \\
\Rightarrow\left\{\begin{array} { l } 
{ a _ { n + 1 } = a _ { n } + 1 , } \\
{ b _ { n + 1 } = - b _ { n } + n }
\end{array} \text { and } \left\{\begin{array}{l}
a_{1}=1, \\
b_{1}=0 .
\end{array} \Rightarrow a_{n}=n\left(n \in \mathbf{N}^{*}\right) .\right.\right. \\
\text { and } b_{n+1}-\frac{1}{2}(n+1)+\frac{1}{4}=-b_{n}+n-\frac{1}{2}(n+1)+\frac{1}{4}=-b_{n}+\frac{1}{2} n-\frac{1}{4} \\
=-\left(b_{n}-\frac{1}{2} n+\frac{1}{4}\right) \Rightarrow b_{n}-\frac{1}{2} n+\frac{1}{4}=-\frac{1}{4}(-1)^{n-1}=\frac{1}{4}(-1)^{n} \\
\Rightarrow b_{n}=\frac{1}{2} n+\frac{1}{4}\left[(-1)^{n}-1\right]\left(n \in \mathbf{N}^{*}\right) .
\end{array}
$$

Therefore, $a_{2015}=2015, b_{2015}=\frac{2015}{2}-\frac{1}{2}=1007 \Rightarrow z_{2015}=2015+1007 \mathrm{i}$.