Example 1 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 the value of $\tan (\alpha+$
$\beta$ ).
From the given, we have
$$
\begin{aligned}
z+u & =(\cos \alpha+\mathrm{i} \sin \alpha)+(\cos \beta+\mathrm{i} \sin \beta)=(\cos \alpha+\cos \beta)+\mathrm{i}(\sin \alpha+\sin \beta) \\
& =\frac{4}{5}+\frac{3}{5} \mathrm{i} .
\end{aligned}
$$

According to the necessary and sufficient condition for the equality of complex numbers, we get
$$
\left\{\begin{array}{l}
\cos \alpha+\cos \beta=\frac{4}{5} \\
\sin \alpha+\sin \beta=\frac{3}{5} .
\end{array}\right.
$$
(1) and (2) are transformed using sum-to-product identities, yielding
$$
\left\{\begin{array}{l}
2 \cos \frac{\alpha+\beta}{2} \cos \frac{\alpha-\beta}{2}=\frac{4}{5}, \\
2 \sin \frac{\alpha+\beta}{2} \cos \frac{\alpha-\beta}{2}=\frac{3}{5} .
\end{array}\right.
$$
Dividing (4) by (3), we get $\tan \frac{\alpha+\beta}{2}=\frac{3}{4}$.

Therefore, $\tan (\alpha+\beta)=\frac{2 \tan \frac{\alpha+\beta}{2}}{1-\tan ^{2} \frac{\alpha+\beta}{2}}=\frac{2 \times \frac{3}{4}}{1-\left(\frac{3}{4}\right)^{2}}=\frac{4}{7}$.