Example 4 Given the real-coefficient equation $x^{3}+2(k-1) x^{2}+9 x+5(k-1)=0$ has a complex root with modulus $\sqrt{5}$, find the value of $k$, and solve this equation.



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Example 4 Given the real-coefficient equation $x^{3}+2(k-1) x^{2}+9 x+5(k-1)=0$ has a complex root with modulus $\sqrt{5}$, find the value of $k$, and solve this equation.
Analysis: By the theorem of complex conjugate roots for real coefficient equations, we know that the original equation has one real root and two conjugate complex roots with a modulus of $\sqrt{5}$. We can solve this using Vieta's formulas.

Solution: By the theorem of complex conjugate roots for real coefficient equations, we know that the original equation has one real root and two complex roots with a modulus of $\sqrt{5}$, which are conjugate to each other. Let these three roots be $a+b \mathrm{i}, a-b \mathrm{i}, c(a, b, c \in \mathbf{R}, b \neq 0)$, then $a^{2}+b^{2}=5$.
Using Vieta's formulas, we get $\left\{\begin{array}{l}(a+b \mathrm{i})+(a-b \mathrm{i})+c=-2(k-1) \\ (a+b \mathrm{i})(a-b \mathrm{i})+c(a+b \mathrm{i})+c(a-b \mathrm{i})=9 \\ (a+b \mathrm{i})(a-b \mathrm{i}) c=-5(k-1)\end{array}\right.$
Combining (1), we rearrange to get $\left\{\begin{array}{l}2 a+c=-2(k-1) \\ a c=2 \\ c=-k+1\end{array}\right.$

From (2) and (4), we know $c=1-k, a=\frac{1}{2}(1-k)$, and substituting these into (3), we get $k=-1$ or 3.
Now, we find the solutions to the original equation: When $k=-1$, the solutions to the original equation are $1+2 \mathrm{i}, 1-2 \mathrm{i}, 2$; when $k=3$, the solutions to the original equation are $-1+2 \mathrm{i},-1-2 \mathrm{i},-2$.