Let $\triangle A B C$ have its incircle center at the origin, with radius $r$, touching $B C, C A, A B$ at $D, E, F$, respectively. The complex numbers corresponding to $D, E, F$ are $t_{1}, t_{2}, t_{3}$. Try to express the circumradius $R$ of $\triangle A B C$ using $r, t_{1}, t_{2}, t_{3}$.
From problem 10.3.4, we know that the complex numbers corresponding to $A$, $B$, and $C$ are $\frac{2 t_{2} t_{3}}{t_{2}+t_{3}}$, $\frac{2 t_{1} t_{3}}{t_{1}+t_{3}}$, and $\frac{2 t_{1} t_{2}}{t_{1}+t_{2}}$, respectively. By the formula for the area of a triangle, we have
$$
A B \cdot B C \cdot C A=4 R \cdot S_{\triangle A B C}, \quad D E \cdot E F \cdot F D=4 r \cdot S_{\triangle D E F},
$$

Thus,
$$
\frac{S_{\triangle D E F}}{S_{\triangle A B C}}-\frac{R}{r} \cdot \frac{D E \cdot E F \cdot F D}{A B \cdot B C \cdot C A}
$$

Also,
$$
\begin{aligned}
A B & =\left|\frac{2 t_{2} t_{3}}{t_{2}+t_{3}}-\frac{2 t_{1} t_{3}}{t_{1}+t_{3}}\right|=2\left|t_{3}\right|^{2}\left|\frac{t_{2}-t_{1}}{\left(t_{1}+t_{3}\right)\left(t_{2}+t_{3}\right)}\right| \\
& =\frac{2 r^{2}}{\left|\left(t_{1}+t_{3}\right)\left(t_{2}+t_{3}\right)\right|} \cdot D E,
\end{aligned}
$$

Thus, $\frac{D E}{A B}=\frac{\left|\left(t_{1}+t_{3}\right)\left(t_{2}+t_{3}\right)\right|}{2 r^{2}}$, and similarly for the other two expressions, so
$$
\frac{S_{\triangle D E F}}{S_{\triangle A B C}}=\frac{R}{8 r^{7}}\left|\left(t_{1}+t_{2}\right)\left(t_{2}+t_{3}\right)\left(t_{3}+t_{1}\right)\right|^{2}.
$$

Also, $S_{\triangle D E F}=S_{\triangle D I E}+S_{\triangle E I F}+S_{\triangle F I D}=\frac{1}{2} r^{2}(\sin A+\sin B+\sin C)$
$$
=\frac{r^{2}(A B+B C+C D)}{4 R}=\frac{r}{2 R} \cdot S_{\triangle A B C},
$$

Thus, $\frac{R}{8 r^{7}}\left|\left(t_{1}+t_{2}\right)\left(t_{2}+t_{3}\right)\left(t_{3}+t_{1}\right)\right|^{2}=\frac{r}{2 R}$. Therefore, $R=\frac{2 r^{4}}{\left|\left(t_{1}+t_{2}\right)\left(t_{2}+t_{3}\right)\left(t_{3}+t_{1}\right)\right|}$.