//

吴康: 单位根的复合函数值问题

单位根的复合函数值问题

吴康

受文[1][2]的启发, 笔者针对单位根的复合函数求值技巧, 命制两道计算题, 供参考. 题目编号承接[1][2].

题3. 设\omega=\cos\dfrac{2\pi}{5}+\mathrm{i}\sin\dfrac{2\pi}{5}, f\left(x\right)=x^2+x+2, 对

3A. F\left(x\right)=\left(x-2\right)\left(x+1\right),

3B. F\left(x\right)=\left(x^2+1\right)\left(x^2-2x-1\right),

3C. F\left(x\right)=x^3+x+1,

分别求值:

    \[T=F\Big(f\big(\omega\big)\Big)F\Big(f\big(\omega^2\big)\Big)F\Big(f\big(\omega^3\big)\Big)F\Big(f\big(\omega^4\big)\Big). \qquad (1)\]

解: 由单位根的性质, 易知

    \[\prod\limits_{k=0}^{4}\left(x-\omega^k\right)=x^5-1\Longrightarrow \prod\limits_{k=1}^{4}\left(x-\omega^k\right)=x^4+x^3+x^2+x+1. \qquad (2)\]

(2)右端为g\left(x\right), 令f_a\left(x\right)=x^2+x+a, 易算得

    \[g\left(x\right)=\left(x^2+1-a\right)f_a\left(x\right)+ax+1-a+a^2. \qquad (3)\]

f_a\left(x\right)两根为x_1, x_2, 由因式定理和韦达定理知

    \[f_a\left(x\right)=\left(x-x_1\right)\left(x-x_2\right), x_1+x_2=-1, x_1 x_2=a, \qquad (4)\]

故由(3)(4)式得

    \[\begin{aligned}L_a&=\prod\limits_{k=1}^{4}f_a\left(\omega^k\right)=\prod\limits_{k=1}^{4}\prod\limits_{i=1}^{2}\left(\omega^k-x_i\right)=\prod\limits_{i=1}^{2}\prod\limits_{k=1}^{4}\left(x_i-\omega^k\right)\\&=\prod\limits_{i=1}^{2}g\left(x_i\right)=\prod\limits_{i=1}^{2}\left(ax_i+1-a+a^2\right)\\&=a^2x_1 x_2+a\left(1-a+a^2\right)\left(x_1+x_2\right)+\left(1-a+a^2\right)^2\\&=a^3-a\left(1-a+a^2\right)+\left(1-a+a^2\right)^2\\&=a^4-2a^3+4a^2-3a+1. \qquad\qquad\qquad\qquad\qquad\qquad (5)\end{aligned}\]

3A. 对F\left(x\right)=\left(x-2\right)\left(x+1\right), 由(5)式得

    \[\begin{aligned}T&=\prod\limits_{k=1}^{4}F\Big(f\big(\omega^k\big)\Big)=\prod\limits_{k=1}^{4}\prod\limits_{i=1}^{2}\Big[f\big(\omega^k\big)+r\Big]\\&=\prod\limits_{r=-2, 1}\prod\limits_{k=1}^{4}f_{2+r}\big(\omega^k\big)=\prod\limits_{r=-2, 1}L_{2+r}\\&=\prod\limits_{a=0, 3}\big(a^4-2a^3+4a^2-3a+1\big)=1\times 55=55. \qquad\qquad (6)\end{aligned}\]

3B. 对F\left(x\right)=\left(x^2+1\right)\left(x^2-2x-1\right), 易见其4根为\alpha_1=\mathrm{i}, \alpha_2=-\mathrm{i}, \alpha_3=1+\sqrt{2}, \alpha_4=1-\sqrt{2}, 故F\left(x\right)=\prod\limits_{r=1}^{4}\big(x-\alpha_r\big), 且

    \[\begin{aligned}T&=\prod\limits_{k=1}^{4}F\Big(f\big(\omega^k\big)\Big)=\prod\limits_{k=1}^{4}\prod\limits_{r=1}^{4}\Big[f\big(\omega^k\big)-\alpha_r\Big]=\prod\limits_{r=1}^{4}\prod\limits_{k=1}^{4}f_{2-\alpha_r}\big(\omega^k\big)\\&=\prod\limits_{r=1}^{4}L_{2-\alpha_r}=\prod\limits_{a=2\pm \mathrm{i}, 1\pm \sqrt{2}} L_a. \qquad\qquad (7)\end{aligned}\]

易见2\pm\mathrm{i}h_1\left(x\right)=x^2-4x+5的两根, 1\pm \sqrt{2}h_2\left(x\right)=x^2-2x-1的两根, 且

    \[L_x=\left(x^2+2x+7\right)h_1\left(x\right)+15x-34=\left(x^2+5\right)h_2\left(x\right)+7x+6, \qquad (8)\]

故由(7)(8)式可得

    \[\begin{aligned}&T_1=\prod\limits_{a=2\pm\mathrm{i}}L_a=\prod\limits_{a=2\pm\mathrm{i}}\left(15a-34\right)=\left(-4+15\mathrm{i}\right)\left(-4-15\mathrm{i}\right)=241, \\&T_2=\prod\limits_{a=1\pm\sqrt{2}}L_a=\prod\limits_{a=1\pm\sqrt{2}}\left(7a+6\right)=\left(13+7\sqrt{2}\right)\left(13-7\sqrt{2}\right)=-29, \\\Longrightarrow &\,T=T_1 T_2=241\times \left(-29\right)=-6989. \qquad\qquad \qquad\qquad \qquad\qquad (9)\end{aligned}\]

3C. 对F\left(x\right)=x^3+x+1, 设其3根为\alpha_1, \alpha_2, \alpha_3, 故F\left(x\right)=\prod\limits_{r=1}^{3}\left(x-\alpha_r\right), 且

    \[\begin{aligned}T&=\prod\limits_{k=1}^{4}F\Big(f\big(\omega^k\big)\Big)=\prod\limits_{k=1}^{4}\prod\limits_{r=1}^{3}\Big[f\big(\omega^k\big)-\alpha_r\Big]=\prod\limits_{r=1}^{3}\prod\limits_{k=1}^{4}f_{2-\alpha_r}\big(\omega^k\big)\\&=\prod\limits_{r=1}^{3}L_{2-\alpha_r}. \qquad\qquad\qquad\qquad\qquad\qquad (10)\end{aligned}\]

易见\beta_r=2-\alpha_r\left(r=1, 2, 3\right)G\left(x\right)=-F\left(2-x\right)=x^3-6x^2+13x-11的3根, 且G\left(x\right)=\prod\limits_{r=1}^{3}\left(x-\beta_r\right), L_x=\left(x+1\right)G\left(x\right)+h\left(x\right), 其中h\left(x\right)=15x^2-44x+45. 设h\left(x\right)两根为\gamma_1, \gamma_2, 则h\left(x\right)=15\prod\limits_{s=1}^{2}\left(x-\gamma_s\right), 且

待续