Example 2. In $\triangle \mathrm{ABC}$, the length of $\mathrm{AB}$ is $8, \angle C=45^{\circ}$. If the position of $\mathrm{AB}$ remains unchanged, find the equation of the locus of vertex C.
Solve As shown in Figure 3, take the line AB as the x-axis, and the perpendicular bisector of AB as the y-axis, so A(-4,0), B(4, 0). Let the coordinates of the moving point C be (x, y).
i) If point C is above the x-axis, $\angle \mathrm{CAB}+45^{\circ}=\angle \mathrm{CBX}$.
$$
\begin{array}{l}
\because \operatorname{tg} \angle \mathrm{CBX}=\frac{1+\operatorname{tg} \angle \mathrm{CAB}}{1-\operatorname{tg} \angle \mathrm{CAB}}, \\
\therefore \frac{1+\frac{y}{x-4}}{1-\frac{y}{x+4}}=\frac{y}{x-4} .
\end{array}
$$

Simplifying, we get: $x^{2}+(y-4)^{2}=32$.
ii) If C is below the x-axis, we get $x^{2}+(y+4)^{2}=32$.
Therefore, the trajectory equation of point C is: the upper part of $x^{2}+(y-4)^{2}=32$, and the lower part of $x^{2}+(y+4)^{2}=32$.