5. The complex numbers $z_{1}, z_{2}, \cdots, z_{100}$ satisfy: $z_{1}=3+2 \mathrm{i}, z_{n+1}=\overline{z_{n}} \cdot \mathrm{i}^{n}(n=1,2, \cdots, 99)(\mathrm{i}$ is the imaginary unit), then the value of $z_{99}+z_{100}$ is $\qquad$
$$
\begin{array}{l}
z_{1}=3+2 \mathrm{i}, z_{2}=(3-2 \mathrm{i}) \cdot \mathrm{i}=2+3 \mathrm{i}, z_{3}=(2-3 \mathrm{i}) \cdot \mathrm{i}^{2}=-2+3 \mathrm{i}, z_{4}=(-2-3 \mathrm{i}) \cdot \mathrm{i}^{3} \\
=-3+2 \mathrm{i}, z_{5}=(-3-2 \mathrm{i}) \cdot \mathrm{i}^{4}=-3-2 \mathrm{i}, z_{6}=(-3+2 \mathrm{i}) \cdot \mathrm{i}^{5}=-2-3 \mathrm{i}, \\
z_{7}=(-2+3 \mathrm{i}) \cdot \mathrm{i}^{6}=2-3 \mathrm{i}, z_{8}=(2+3 \mathrm{i}) \cdot \mathrm{i}^{7}=3-2 \mathrm{i}, z_{9}=(3+2 \mathrm{i}) \cdot \mathrm{i}^{8}=3+2 \mathrm{i},
\end{array}
$$

That is, $\left\{z_{n}\right\}$ is a periodic sequence with a period of 8, so $z_{99}+z_{100}=z_{3}+z_{4}=-5+5 \mathrm{i}$.