題目：\(x^4-2x^3+6x^2+6x-1=0\) 的四個根在高斯平面表為 \(A,B,C,D\) 四點，若複數 \(i\) 表為 \(P\) 點,求 \(\overline{PA}\times \overline{PB}\times \overline{PC}\times\overline{PD}\)=?
解答：
設 \(x^4-2x^3+6x^2+6x-1=0\) 的四個根分別為 \(\alpha,\beta,\gamma,\delta\)，則
\(\left(x-\alpha\right)\left(x-\beta\right)\left(x-\gamma\right)\left(x-\delta\right)=x^4-2x^3+6x^2+6x-1\)
所求 \(\overline{PA}\times \overline{PB}\times \overline{PC}\times\overline{PD}=\left|i-\alpha\right|\times\left|i-\beta\right|\times\left|i-\gamma\right|\times\left|i-\delta\right|\)
\(=\left|\left(i-\alpha\right)\left(i-\beta\right)\left(i-\gamma\right)\left(i-\delta\right)\right|\)
\(=\left|i^4-2i^3+6i^2+6i-1\right|\)
\(=\left|-6+8i\right|\)
\(=10.\)
