5.3.5 ** The arguments of the complex numbers $z_{1}, z_{2}, z_{3}$ are $\alpha, \beta, \gamma$, respectively, and $\left|z_{1}\right|=1,\left|z_{2}\right|=k$, $\left|z_{3}\right|=2-k, z_{1}+z_{2}+z_{3}=0$. For what value of $k$ does $\cos (\beta-\gamma)$ attain its maximum or minimum value, and what are these maximum and minimum values?
Given $z_{1}+z_{2}+z_{3}=0$, the sum of the corresponding 3 vectors is the zero vector. As shown in the figure, we have
$$
\left\{\begin{array}{l}
1+k \geqslant 2-k, \\
(2-k)+1 \geqslant k, \\
(2-k)+k \geqslant 1,
\end{array}\right.
$$

Solving this, we get
$$
\frac{1}{2} \leqslant k \leqslant \frac{3}{2} \text {. }
$$

According to the cosine rule, $1^{2}=(2-k)^{2}+k^{2}-2(2-k) k \cos [\pi-(\beta-\gamma)]$, so
$$
\cos (\beta-\gamma)=\frac{-3+4 k-2 k^{2}}{4 k-2 k^{2}}=1+\frac{3}{2(k-1)^{2}-2} .
$$

From (1) and (2), we know that when $k=1$, $\cos (\beta-\gamma)$ reaches its maximum value of $-\frac{1}{2}$; when $k=\frac{1}{2}, \frac{3}{2}$, $\cos (\beta-\gamma)$ reaches its minimum value of -1.