高斯代数基本定理的一种证明

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代数基本定理

对于多项式 f(z)=anzn+an1zn1++a1z+a0f(z) = a_n z^n + a_{n-1} z^{n-1} + \cdots + a_1 z + a_0(其中 n>1n > 1an,a00a_n, a_0 \neq 0),它在复数域内有根。

f(z)=U(r,θ)+V(r,θ)if(z) = U(r, \theta) + V(r, \theta)i

其中 rrθ\theta 分别是 zz 的模和幅角。

CR条件

rUr=VθrVr=Uθ\begin{align*} r\frac{\partial U}{\partial r} &= \frac{\partial V}{\partial \theta} \\ r\frac{\partial V}{\partial r} &= -\frac{\partial U}{\partial \theta} \end{align*}

构造二元实函数 HH

H(r,θ)=arctan(U(r,θ)V(r,θ))H(r, \theta) = \arctan\left(\frac{U(r, \theta)}{V(r, \theta)}\right)

HH 的二阶混合偏导数

2Hrθ=r(UθVUVθV2+U2)\frac{\partial^2 H}{\partial r \partial \theta} = \frac{\partial}{\partial r} \left( \frac{\frac{\partial U}{\partial \theta}V - U\frac{\partial V}{\partial \theta}}{V^2 + U^2} \right)

累次积分 I1I_1I2I_2

I1=0Rdr02π2Hrθdθ,I2=02πdθ0R2HrθdrI_1 = \int_{0}^{R} dr \int_{0}^{2\pi} \frac{\partial^2 H}{\partial r \partial \theta} d\theta, \quad I_2 = \int_{0}^{2\pi} d\theta \int_{0}^{R} \frac{\partial^2 H}{\partial r \partial \theta} dr

假设 f(z)f(z) 无根

假设 f(z)f(z) 无根,则 U2+V20U^2 + V^2 \neq 0,从而 2Hrθ\frac{\partial^2 H}{\partial r \partial \theta} 连续,积分顺序可交换,并且 I1=I2I_1 = I_2

I1I_1I2I_2 不恒等

02π2Hrθdθ=0    I1=0\int_{0}^{2\pi} \frac{\partial^2 H}{\partial r \partial \theta} d\theta = 0 \implies I_1 = 0
I2=limR02πdθ0R2Hrθdr=02πdθ(Hθr=0r=R)=2πnI_2 = \lim_{R \to \infty} \int_{0}^{2\pi} d\theta \int_{0}^{R} \frac{\partial^2 H}{\partial r \partial \theta} dr = \int_{0}^{2\pi} d\theta \left( \left. \frac{\partial H}{\partial \theta} \right|_{r=0}^{r=R} \right) = -2\pi n

结论

由于 I1I2I_1 \neq I_2,存在 rm,θmr_m, \theta_m 使得 U2(rm,θm)+V2(rm,θm)=0U^2(r_m, \theta_m) + V^2(r_m, \theta_m) = 0,即存在某个 zm=rmeiθmz_m = r_m e^{i\theta_m} 使得 f(zm)=0f(z_m) = 0