代数基本定理
对于多项式 f(z)=anzn+an−1zn−1+⋯+a1z+a0f(z) = a_n z^n + a_{n-1} z^{n-1} + \cdots + a_1 z + a_0f(z)=anzn+an−1zn−1+⋯+a1z+a0（其中 n>1n > 1n>1 且 an,a0≠0a_n, a_0 \neq 0an,a0=0），它在复数域内有根。
f(z)=U(r,θ)+V(r,θ)i f(z) = U(r, \theta) + V(r, \theta)i f(z)=U(r,θ)+V(r,θ)i
其中 rrr 和 θ\thetaθ 分别是 zzz 的模和幅角。
CR条件
r∂U∂r=∂V∂θr∂V∂r=−∂U∂θ \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*} r∂r∂Ur∂r∂V=∂θ∂V=−∂θ∂U
构造二元实函数 HHH
H(r,θ)=arctan(U(r,θ)V(r,θ)) H(r, \theta) = \arctan\left(\frac{U(r, \theta)}{V(r, \theta)}\right) H(r,θ)=arctan(V(r,θ)U(r,θ))
HHH 的二阶混合偏导数
∂2H∂r∂θ=∂∂r(∂U∂θV−U∂V∂θV2+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) ∂r∂θ∂2H=∂r∂(V2+U2∂θ∂UV−U∂θ∂V)
累次积分 I1I_1I1 和 I2I_2I2
I1=∫0Rdr∫02π∂2H∂r∂θdθ,I2=∫02πdθ∫0R∂2H∂r∂θdr I_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 I1=∫0Rdr∫02π∂r∂θ∂2Hdθ,I2=∫02πdθ∫0R∂r∂θ∂2Hdr
假设 f(z)f(z)f(z) 无根
假设 f(z)f(z)f(z) 无根，则 U2+V2≠0U^2 + V^2 \neq 0U2+V2=0，从而 ∂2H∂r∂θ\frac{\partial^2 H}{\partial r \partial \theta}∂r∂θ∂2H 连续，积分顺序可交换，并且 I1=I2I_1 = I_2I1=I2。
I1I_1I1 和 I2I_2I2 不恒等
∫02π∂2H∂r∂θdθ=0 ⟹ I1=0 \int_{0}^{2\pi} \frac{\partial^2 H}{\partial r \partial \theta} d\theta = 0 \implies I_1 = 0 ∫02π∂r∂θ∂2Hdθ=0⟹I1=0
I2=limR→∞∫02πdθ∫0R∂2H∂r∂θdr=∫02πdθ(∂H∂θ∣r=0r=R)=−2πn I_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 I2=R→∞lim∫02πdθ∫0R∂r∂θ∂2Hdr=∫02πdθ(∂θ∂Hr=0r=R)=−2πn
结论
由于 I1≠I2I_1 \neq I_2I1=I2，存在 rm,θmr_m, \theta_mrm,θm 使得 U2(rm,θm)+V2(rm,θm)=0U^2(r_m, \theta_m) + V^2(r_m, \theta_m) = 0U2(rm,θm)+V2(rm,θm)=0，即存在某个 zm=rmeiθmz_m = r_m e^{i\theta_m}zm=rmeiθm 使得 f(zm)=0f(z_m) = 0f(zm)=0。
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