Question 103: Let $z, w$ be complex numbers satisfying $\left\{\begin{array}{c}z+\frac{20 i}{w}=5+i \\ w+\frac{12 i}{z}=-4+10 i\end{array}\right.$, find the minimum possible value of $|z w|^{2}$.
Question 103, Solution: Multiplying the two expressions yields $\mathrm{zw}-\frac{240}{z \mathrm{w}}+12 \mathrm{i}+20 \mathrm{i}=(5+\mathrm{i})(-4+10 \mathrm{i})=-30+$ $46 \mathrm{i}$, rearranging gives
$$
\begin{array}{l}
(\mathrm{zw})^{2}-(-30+14 \mathrm{i}) \cdot \mathrm{zw}-240=0 \\
\Rightarrow[\mathrm{zw}-(6+2 \mathrm{i})][\mathrm{zw}-(-36+12 \mathrm{i})]=0
\end{array}
$$

Thus, $\mathrm{zw}=6+2 \mathrm{i}$ or $-36+12 \mathrm{i}$. Therefore, the minimum possible value of $|\mathrm{zw}|^{2}$ is $6^{2}+2^{2}=40$.