Example 1. Let $z$ be a complex number, solve the equation
$$
\frac{1}{2}(z-1)=\frac{\sqrt{3}}{2}(1+z) \text { i. }
$$
Solving by eliminating $\frac{1}{2}$ from both sides, we get
$$
z-1=\sqrt{3} i+\sqrt{3} i \cdot z .
$$

Rearranging and combining terms, we get
$$
(1 -\sqrt{3} i) z=1+\sqrt{3} i \text {. }
$$

This leads to $z=\frac{1+\sqrt{3}}{1-\sqrt{3}} \frac{i}{i}=-\frac{1}{2}+\frac{\sqrt{3}}{2} i$.
Or, let $z=a+b i,(a, b \in R)$
From $\frac{1}{2}(a+b i-1)=\frac{\sqrt{3}}{2}(1+a+b i) i$, we get
$$
\left(\frac{1}{2} a-\frac{1}{2}\right)+\frac{1}{2} b i=-\frac{\sqrt{3}}{2} b+\left(\frac{\sqrt{3}}{2}+\frac{\sqrt{3}}{2} a\right) i \text {. }
$$

By the condition of equality of complex numbers, we get
$$
\left\{\begin{array}{l}
\frac{1}{2} a-\frac{1}{2}=-\frac{\sqrt{3}}{2} b, \\
\frac{\sqrt{3}}{2}+\frac{\sqrt{3}}{2} a=\frac{1}{2} b .
\end{array}\right.
$$

Solving this system of equations, we get $a=-\frac{1}{2}, \quad b=\frac{\sqrt{3}}{2}$.
Therefore, $z=-\frac{1}{2}+\frac{\sqrt{3}}{2} i$.
This solution method aligns with the general thought process of students and reveals a general approach to solving such problems.
We can transform the original equation into
$$
-\frac{1}{2}-\frac{\sqrt{3}}{2} i=\left(-\frac{1}{2}+\frac{\sqrt{3}}{2} i\right) z \text {. }
$$

Thus, we can derive $z=-\frac{1}{2}+\frac{\sqrt{3}}{2} i$. Because we know
$$
-\frac{1}{2}+\frac{\sqrt{3}}{2} i, \quad-\frac{1}{2}-\frac{\sqrt{3}}{2} i
$$

are the two cube roots of 1, $\omega$ and $\omega^{2}$. The equation essentially solves $\omega^{2}=\omega \cdot z$.
This method is simple and clear, reflecting the editor's intention. Here,