13. In $\triangle A B C$, $B C=a, C A=b$, a square $A B E F$ is constructed outward with side $A B$, $O$ is the center of the square $A B E F$, $M$ and $N$ are the midpoints of $B C$ and $C A$ respectively. When $\angle B C A$ varies, find the maximum value of $O M+O N$
13. Solution: As shown in the figure, connect $C F, C E$, then $O M=\frac{1}{2} C F, O N$ $=\frac{1}{2} C E$. Establish a complex plane with $C$ as the origin, and use the names of each point to represent the corresponding complex numbers,
$$
\begin{array}{l}
\quad \text { then } F-A=(B-A) \mathrm{i}, E-B=-(A-B) \mathrm{i}, F=A \\
+(B-A) \mathrm{i}=(1-\mathrm{i}) A+B \mathrm{i}, E=B-(A-B) \mathrm{i}=-A \mathrm{i} \\
+(1+B) \mathrm{i},
\end{array}
$$

Thus, $|F|+|E|=|(1-\mathrm{i}) A+B \mathrm{i}|+\mid-A \mathrm{i}+(1+\mathrm{i})$
$$
\begin{array}{l}
B|\leqslant| 1-\mathrm{i}|\times| A|+| B|+| A|+| 1+\mathrm{i}|\times| B \mid=(\sqrt{2} \\
+1)(a+b),
\end{array}
$$

When the vectors $(1-\mathrm{i}) A \parallel B \mathrm{i},-A \mathrm{i} \parallel (1+\mathrm{i}) B$, the equality holds, at this time, $\angle B C A=135^{\circ},(O M+O N)_{\max }=\frac{\sqrt{2}+1}{2}(a+b)$.