Example: Let $P_{1}, P_{2}, \cdots, P_{n}$ be any $n$ points given in the plane, try to find a point $A$ such that $A P_{1}^{2}+A P_{2}^{2}+\cdots+A P_{n}^{2}$ is minimized.
Analysis Using the distance formula between two points in the complex plane and
$$
\left|z_{1}-z_{2}\right|^{2}=\left(z_{1}-z_{2}\right)\left(\bar{z}_{1}-\bar{z}_{2}\right)=\left|z_{1}\right|^{2}+\left|z_{2}\right|^{2}-\left(z_{1} \bar{z}_{2}+\bar{z}_{1} z_{2}\right) .
$$

Solution Let the complex numbers corresponding to $P_{1}, P_{2}, \cdots, P_{n}, A$ be $z_{1}, z_{2}, \cdots, z_{n}, z$, respectively, then
$$
\begin{aligned}
A P_{1}{ }^{2} & +A P_{2}{ }^{2}+\cdots+A P_{n}{ }^{2}=\sum_{k=1}^{n}\left|z-z_{k}\right|^{2}=\sum_{k=1}^{n}\left(z-z_{k}\right)\left(\bar{z}-\bar{z}_{k}\right) \\
& =\sum_{k=1}^{n} z \bar{z}-\left(\sum_{k=1}^{n} z_{k}\right) \bar{z}-\left(\sum_{k=1}^{n} \bar{z}_{k}\right) z+\sum_{k=1}^{n} z_{k} \bar{z}_{k} \\
& =n\left(z-\frac{1}{n} \sum_{k=1}^{n} z_{k}\right)\left(\bar{z}-\frac{1}{n} \sum_{k=1}^{n} \bar{z}_{k}\right)+\sum_{k=1}^{n} z_{k} \bar{z}_{k}-\frac{1}{n}\left(\sum_{k=1}^{n} z_{k}\right)\left(\sum_{k=1}^{n} \bar{z}_{k}\right) \\
& =n\left|z-\frac{1}{n} \sum_{k=1}^{n} z_{k}\right|^{2}+\sum_{k=1}^{n}\left|z_{k}\right|^{2}-\frac{1}{n}\left|\sum_{k=1}^{n} z_{k}\right|^{2} \geqslant \sum_{k=1}^{n}\left|z_{k}\right|^{2}-\frac{1}{n}\left|\sum_{k=1}^{n} z_{k}\right|^{2} .
\end{aligned}
$$

Equality holds if and only if $z=\frac{1}{n} \sum_{k=1}^{n} z_{k}$. Therefore, when $z=\frac{1}{n} \sum_{k=1}^{n} z_{k}$, $A P_{1}{ }^{2}+A P_{2}{ }^{2}+\cdots+A P_{n}{ }^{2}$ achieves its minimum value $\sum_{k=1}^{n}\left|z_{k}\right|^{2}-\frac{1}{n}\left|\sum_{k=1}^{n} z_{k}\right|^{2}$.