Mar 232013

Fundamental theorem of algebra(FTA)   Every polynomial of degree \(n\geqslant1\) with complex coeficients has a zero in \(\Bbb C\).

我们尝试使用 Cauchy 积分公式来证明代数基本定理. 其实有几个大同小异的证明, 基本的想法是一致的. 第一个证明属于 Anton R. Schep.

Proof. Let \(p(z)=z^n+a_{n-1}z^{n-1}+\dotsb+a_1z+a_0\) be a polynomial of degree \(n\geqslant1\) and assume that \(p(z)\ne0\) for all \(z\in\Bbb C\). Then the function defined by \(\frac1{p(z)}\) is entire. By Cauchy’s theorem,

\[\int_{|z|=r}\dfrac{\mathrm dz}{zp(z)}=\dfrac{2\,\mathrm\pi i}{p(0)}\ne 0.\]


\[|p(z)| \geqslant |z|^n\left(1-\frac{|a_{n-1}|}{|z|}-\dotsb-\frac{|a_0|}{|z|^n}\right) > \frac12|z|^n\]

for \(z\) sufficiently large, so

\[\left|\int_{|z|=r}\dfrac{\mathrm dz}{zp(z)}\right| \leqslant 2\,\mathrm\pi r \cdot \max_{|z|=r}\dfrac1{|zp(z)|} = \dfrac{2\,\mathrm\pi }{\min\limits_{|z|=r}|p(z)|}\to0(r\to\infty),\]

which is a contradiction, and therefore \(p(z)\) has a zero.   \(\Box\)

这个证明还可以另一种面目表现出来: 使用关于解析函数的 Mean-Value Property(MVP).

第二个证明   依旧是反证法. 我们可以假定 \(z\in\Bbb R\) 时, \(p(z)\) 是实数(否则, 考虑 \(p(z)\overline{p}(z)\), 这里 \(\overline{p}(z)=z^n+\overline{a_{n-1}}z^{n-1}+\dotsb+\overline{a_1}z+\overline{a_0}\)). 既然, \(\forall x\in\Bbb R, p(x)\ne0\), 于是可有

\[\int_0^{2\,\mathrm\pi}\dfrac{\mathrm d\theta}{p(2\cos \theta)}\ne 0.\]

但是, 这个积分显然等于

\[\frac1i\int_{|z|=1}\frac{\mathrm dz}{zp(z+\frac1z)}=\frac1i\int_{|z|=1}\frac{z^{n-1}\,\mathrm dz}{q(z)},\]

这里 \(q(z)=z^np(z+\frac1z)\) 是一个多项式. 显然 \(z\ne0\) 时, \(q(z)\ne 0\), 而且 \(q(0)=1\), 于是 \(\dfrac{z^{n-1}}{q(z)}\) 是整函数. 据 Cauchy 定理, 上面的积分是 \(0\). 矛盾!

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