Example 2. Consider a parallelogram $O A B C$, where point $O$ is fixed at the origin, point $A$ moves on the line segment $x=1$, $y=t \quad(-1 \leqslant t \leqslant 1)$, and point $C$ moves on the unit circle. Find the equation of the trajectory of point $B$.
Solution 1: Let the complex numbers corresponding to points $A, B, C$ be $1+t i(-1 \leqslant t \leqslant 1), x+y i, \cos \theta +i \sin \theta . \quad(0 \leqslant 0 \leqslant 2 \pi)$
$\because O A B C$ is a parallelogram,
$$
\begin{array}{c}
\therefore \overrightarrow{O B}=\overrightarrow{O A}+ \\
\overrightarrow{A B}=\overrightarrow{O A}+\overrightarrow{O C} .
\end{array}
$$
$$
\begin{array}{l}
\text { i.e., } x+y i=(1+t i) \\
+(\cos \theta+i \sin \theta) \\
=(1+\cos \theta) \\
+(t+\sin \theta) i,
\end{array}
$$

From the condition of equality of complex numbers, we get: $\left\{\begin{array}{l}x=1+\cos \theta \\ y=t+\sin \theta\end{array}\right.$
Eliminating the parameter $\theta$ yields:
$$
(x-1)^{2}+(y-t)^{2}=1 . \quad(-1 \leqslant t \leqslant 1)
$$

That is, the trajectory of point $B$ is a circle with center $(1, t)$ and radius 1 (obviously $t$ can vary in $[-1,1]$, these trajectories form a region consisting of a square with side length 2 plus several semicircles).
Solution 2: Set as in Solution 1.
$$
\begin{array}{l} 
\because A B=\overrightarrow{O C}, \\
\therefore(x-1)+(y-t) i=\cos \theta+\sin \theta . \\
\Rightarrow\left\{\begin{array}{l}
x=1+\cos \theta, \\
y=t+\sin \theta
\end{array} \Rightarrow(x-1)^{2}+(y-t)^{2}=1\right.
\end{array}
$$

This is the trajectory of point $B$ $(-1 \leqslant t \leqslant 1)$.
Solution 3: Let the complex numbers corresponding to points $A, B, C$ be $1+t i(-1 \leqslant t \leqslant 1), x+y i, x_{0}+y_{0} i$
$$
\begin{aligned}
\because \overrightarrow{O C} & =\overrightarrow{O B}-\overrightarrow{O A} \\
& =(x+y i)-(1+t i) \\
& \therefore(x-1)+(y-t) i,
\end{aligned}
$$
$$
\therefore x_{0}=x-1, y_{0}=y-t_{\text {。 }}
$$
$\because C$ point is on the circle,
$$
\therefore x_{0}^{2}+y_{0}^{2}=1 \text {. }
$$

That is, $(x-1)^{2}+(y-t)^{2}=1$.
Example 2 explanation: (Solution 3), the full utilization of the equality relationship of vectors (Solution 2), and the multiple relationship of vectors, brings convenience to problem solving.
$2^{\circ}$ When the problem does not specify the positive or negative value of the rotation angle $\alpha$, it should consider using $\pm \alpha$ to solve the problem to prevent missing solutions.