For a natural number $n$, let $S_{n}$ be
$$
\sum_{k=1}^{n} \sqrt{(2 k-1)^{2}+a_{k}^{2}}
$$

the minimum value, where $a_{1}, a_{2}, \cdots, a_{n}$ are positive real numbers, and their sum is 17. If there exists a unique $n$ such that $S_{n}$ is also an integer, find $n .{ }^{(4]}$
Solve: Regarding $\sum_{k=1}^{n} \sqrt{(2 k-1)^{2}+a_{k}^{2}}$ as the modulus of the complex number $(2 k-1)+a_{k} \mathrm{i}$.
$$
\begin{array}{l}
\text { Hence } \sum_{k=1}^{n} \sqrt{(2 k-1)^{2}+a_{k}^{2}} \\
=\sum_{k=1}^{n}\left|(2 k-1)+a_{k} \mathrm{i}\right| \\
\geqslant\left|\sum_{k=1}^{n}\left[(2 k-1)+a_{k} \mathrm{i}\right]\right| \\
=\left|n^{2}+17 \mathrm{i}\right|=\sqrt{n^{4}+17^{2}}
\end{array}
$$

From the given condition, we have
$$
\begin{array}{l}
n^{4}+17^{2}=m^{2}\left(m \in \mathbf{N}_{+}, m=S_{n}\right) \\
\Rightarrow\left(m-n^{2}\right)\left(m+n^{2}\right)=289 \\
\Rightarrow m-n^{2}=1, m+n^{2}=289 \\
\Rightarrow n=12 .
\end{array}
$$
2.2 Application in Trigonometric Problems

The connection between complex numbers and trigonometric functions mainly relies on the exponential form (or trigonometric form) of complex numbers. By using the exponential form and operations (or trigonometric operations) of complex numbers, we can achieve the purpose of trigonometric evaluation and proof.