Example $2 . \square O A B C$ has its vertex $O$ fixed at the origin, $A$ moves on the line segment $x=1, y=t(-1 \leqslant t \leqslant 1)$, and $C$ moves on the unit circle $\odot O$. Find the trajectory of $B$.
Let $A, B, C$ points correspond to the complex numbers $1+t i$
$$
\begin{array}{l}
(-1 \leqslant t \leqslant 1), x+y i, \cos \theta+i \sin \theta(0 \\
\leqslant\theta \leqslant 2 \pi) \\
\vec{O} \quad \overrightarrow{O C}, \text { for } \square O A B C, \overrightarrow{O B}=\overrightarrow{O A}+\overrightarrow{A B} \\
x+y i=(1+t i)+(\cos \theta+i \sin \theta) \\
=(1+\cos \theta)+i(t+\sin \theta) .
\end{array}
$$

By the condition of equality of complex numbers: $x=1+\cos \theta$, $y=t+\sin \theta$, eliminating $\theta$ :
$$
\begin{array}{l}
(x-1)^{2}+(y-t)^{2}=1 \cdot(-1 \leqslant t \\
\leqslant 1)
\end{array}
$$