51.6 ** Given the complex numbers $z=\cos \alpha+\mathrm{i} \sin \alpha, u=\cos \beta+\mathrm{i} \sin \beta$, and $z+u=\frac{4}{5}+\frac{3}{5} \mathrm{i}$.
(1) Find the value of $\tan (\alpha+\beta)$; (2) Prove that $z^{2}+u^{2}+z u=0$.
(1) According to the problem, we have
$$
\begin{aligned}
z+u & =(\cos \alpha+\mathrm{i} \sin \alpha)+(\cos \beta+\mathrm{i} \sin \beta)-(\cos \alpha+\cos \beta)+\mathrm{i}(\sin \alpha+\sin \beta) \\
& =\frac{4}{5}+\frac{3}{5} \mathrm{i} .
\end{aligned}
$$

By the necessary and sufficient condition for the equality of complex numbers, we know that $\left\{\begin{array}{l}\cos \alpha+\cos \beta=\frac{4}{5}, \\ \sin \alpha+\sin \beta=\frac{3}{5},\end{array}\right.$ Therefore,
$$
\left\{\begin{array}{l}
2 \cos \frac{\alpha+\beta}{2} \cos \frac{\alpha-\beta}{2}=\frac{4}{5}, \\
2 \sin \frac{\alpha+\beta}{2} \cos \frac{\alpha-\beta}{2}=\frac{3}{5} .
\end{array}\right.
$$
Dividing (2) by (1) gives $\tan \frac{\alpha+\beta}{2}=\frac{3}{4}$. Thus, $\tan (\alpha+\beta)=\frac{2 \tan \frac{\alpha+\beta}{2}}{1-\tan ^{2} \frac{\alpha+\beta}{2}}=\frac{24}{7}$.
(2) From (1), we get $\sin (\alpha+\beta)=\frac{2 \tan \frac{\alpha+\beta}{2}}{1+\tan ^{2} \frac{\alpha+\beta}{2}}=\frac{24}{25}$,
$$
\cos (\alpha+\beta)=\frac{1-\tan ^{2} \frac{\alpha+\beta}{2}}{1+\tan ^{2} \frac{\alpha+\beta}{2}}=\frac{7}{25} .
$$

And $z u=(\cos \alpha+\mathrm{i} \sin \alpha)(\cos \beta+\mathrm{i} \sin \beta)=\cos (\alpha+\beta)+\mathrm{i} \sin (\alpha+\beta)=\frac{7}{25}+\frac{24}{25} \mathrm{i}$,
Therefore, $z^{2}+u^{2}+z u=(z+u)^{2}-z u=\left(\frac{4}{5}+\frac{3}{5} \mathrm{i}\right)^{2}-\left(\frac{7}{25}+\frac{24}{25} \mathrm{i}\right)=0$.