Example 1. Given the circle $C: x^{2}+y^{2}=1$ and point $A(2, 0)$, $B$ is a moving point on the circle. Construct an isosceles right triangle $\triangle A B P$ (with $A, B, P$ in clockwise order) using $A B$ as one leg. Try to find the equation of the trajectory of point $P$.
Place circle $C$ and point $\boldsymbol{A}$ on the corresponding complex plane, then the complex number equation of circle $C$ is $|z|=1$. Let $B$ and $P$ correspond to the complex numbers $z^{\prime}$ and $z$, respectively. Clearly, $\left|z^{\prime}\right|=1$.

Vectors $\overrightarrow{B A}$ and $\overrightarrow{B P}$ correspond to the complex numbers $2-z^{\prime}$ and $z-z^{\prime}$, respectively, and $\overrightarrow{B P}$ can be seen as $\overrightarrow{B A}$, with its length unchanged, rotating counterclockwise by $\frac{\pi}{2}$ around point $B$. According to the geometric meaning of multiplication, we have
$$
z-z^{\prime}=i\left(2-z^{\prime}\right) \Rightarrow z^{\prime}=\frac{1}{2}(z-2 i)
$$
- (1+i) . (2)

Taking the modulus on both sides of (2) and substituting into (1),
$$
\begin{array}{l}
\frac{1}{2}|(z-2 i)(1+i)|=1:=1==\sqrt{2} \Rightarrow \\
|z-2 i|=\sqrt{2} .
\end{array}
$$