Example 12 Given that vertex $A$ of the equilateral triangle $\triangle ABC$ is at the vertex of the parabola $y=2x^2$, and point $B$ moves on the parabola. Find the equation of the locus of vertex $C$.

---

The translation maintains the original text's format and line breaks as requested.
Solve: As shown in Figure 9-5, consider the Cartesian coordinate plane as the complex plane, where point $B$ corresponds to the complex number $Z_{B}=x_{1}+y_{1} \mathrm{i}$, and point $C$ corresponds to the complex number $Z_{C}=x+y \mathrm{i}$.

Since $\triangle O B C$ is an equilateral triangle, $\overrightarrow{O C}$ can be obtained by rotating $\overrightarrow{O B}$ counterclockwise or clockwise by $\frac{\pi}{3}$,

Therefore, $x+y \mathrm{i}=\left(x_{1}+\mathrm{i} y_{1}\right)\left[\cos \left( \pm \frac{\pi}{3}\right)+\mathrm{i} \sin \left( \pm \frac{\pi}{3}\right)\right]=$ $\left(x_{1}+\mathrm{i} y_{1}\right)\left(\frac{1}{2} \pm \frac{\sqrt{3}}{2} \mathrm{i}\right)=\left(\frac{1}{2} x_{1} \pm \frac{\sqrt{3}}{2} y_{1}\right)+\left(\frac{1}{2} y_{1} \pm \frac{\sqrt{3}}{2} x_{1}\right) \mathrm{i}$,
By the condition of equality of two complex numbers, we get
$$
\left\{\begin{array} { l } 
{ x = \frac { 1 } { 2 } x _ { 1 } \mp \frac { \sqrt { 3 } } { 2 } y _ { 1 } , } \\
{ y = \frac { 1 } { 2 } y _ { 1 } \pm \frac { \sqrt { 3 } } { 2 } x _ { 1 } . }
\end{array} \text { Therefore, } \left\{\begin{array}{l}
x_{1}=\frac{1}{2} x \pm \frac{\sqrt{3}}{2} y, \\
y_{1}=\frac{1}{2} y \mp \frac{\sqrt{3}}{2} x .
\end{array}\right.\right.
$$

Since point $B$ lies on the parabola, we have $y_{1}=2 x_{1}^{2}$, so $\frac{1}{2} y \mp \frac{\sqrt{3}}{2} x=2\left(\frac{1}{2} x \pm \frac{\sqrt{3}}{2}\right)^{2}$, simplifying gives $x^{2} \pm 2 \sqrt{3} x y+3 y^{2} \pm \sqrt{3} x-y=0$, which is the equation of the locus of vertex $C$.