4. $2016+\mathrm{i}$.

From the problem, we know
$$
\begin{array}{l}
F(F(z))=\frac{\frac{z-\mathrm{i}}{z+\mathrm{i}}-\mathrm{i}}{\frac{z-\mathrm{i}}{z+\mathrm{i}}+\mathrm{i}}=\frac{z-\mathrm{i}-\mathrm{i} z+1}{z-\mathrm{i}+\mathrm{i} z-1} \\
=\frac{1-\mathrm{i}}{1+\mathrm{i}} \cdot \frac{z+1}{z-1}=-\mathrm{i} \cdot \frac{z+1}{z-1} \\
\Rightarrow F(F(F(z)))=F\left(-\mathrm{i} \cdot \frac{z+1}{z-1}\right)=z \\
\Rightarrow z_{n}=z_{n-3}(n \geqslant 3) \\
\Rightarrow z_{2016}=z_{0}=2016+\mathrm{i}
\end{array}
$$
4. For all complex numbers $z$ satisfying $z \neq \mathrm{i}$, we have $F(z) = \frac{z-\mathrm{i}}{z+\mathrm{i}}$. For all positive integers $n$, $z_{n}=F\left(z_{n-1}\right)$. If $z_{0}=2016+\mathrm{i}$, then $z_{2016}=$ $\qquad$ .