14. Given the complex numbers $z=\cos \alpha+\mathrm{i} \sin \alpha, u=\cos \beta+\mathrm{i} \sin \beta$, and $z+u=\frac{4}{5}+\frac{3}{5} \mathrm{i}$, find $\tan (\alpha+\beta)$.
14. Since $z+u=(\cos \alpha+\cos \beta)+\mathrm{i}(\sin \alpha+\sin \beta)=\frac{4}{5}+\frac{3}{5} \mathrm{i}$, we have $\left\{\begin{array}{l}\cos \alpha+\cos \beta=\frac{4}{5}, \\ \sin \alpha+\sin \beta=\frac{3}{5},\end{array}\right.$ which means $\left\{\begin{array}{l}2 \cos \frac{\alpha+\beta}{2} \cos \frac{\alpha-\beta}{2}=\frac{4}{5}, \\ 2 \sin \frac{\alpha+\beta}{2} \cos \frac{\alpha-\beta}{2}=\frac{3}{5},\end{array}\right.$ Therefore, $\tan \frac{\alpha+\beta}{2}=\frac{3}{4}$, hence $\tan (\alpha+\beta)=\frac{24}{7}$.