|
|
lectures14-Earl.pdf page77
Theorem 231 (Fundamental Theorem of Algebra) (Gauss 1799) Let $p$ be a non-constant polynomial with complex coefficients. Then there exists $\alpha \in \mathbb{C}$ such that $p(\alpha)=0$.
Proof. Say that $p(z)=a_n z^n+\cdots+a_0$ where $n \geqslant 1, a_i \in \mathbb{C}$ and $a_n \neq 0$. As $p(z) / z^n \rightarrow a_n$ as $z \rightarrow \infty$ then there exists $R>0$ such that
\[
\left|\frac{p(z)}{z^n}\right|>\frac{\left|a_n\right|}{2} \quad \text { for }|z|>R .\tag{1}
\]Suppose for a contradiction that $p$ has no roots so that $1 / p$ is holomorphic. Then, by Cauchy's Integral Formula, with $r>R$, we have\begin{aligned}
0 & \neq\left|\frac{1}{p(0)}\right|=\left|\frac{1}{2 \pi i} \int_{\gamma(0, r)} \frac{\mathrm{d} w}{w p(w)}\right| \\
& \leqslant \frac{1}{2 \pi} \times 2 \pi r \times \sup _{|w|=r}\left|\frac{1}{w p(w)}\right| \\
& \color{#f00}\leqslant \frac{1}{2 \pi} \times 2 \pi r \times \frac{2}{\left|a_n\right| r^{n+1}} \\
&=\frac{2}{\left|a_n\right| r^n} \rightarrow 0 \quad \text { as } r \rightarrow \infty
\end{aligned}which is the required contradiction.
______________________
Proof of the red inequality:
\[\left|\frac{1}{w p(w)}\right|=\frac1{r\left|p(w)\right|}\overset{(1)}<\frac{2}{\left|a_n\right| r^{n+1}}\quad\text{, for }|w|=r\]
Therefore
\[\sup_{|w|=r}\left|\frac{1}{w p(w)}\right|\leqslant\frac{2}{\left|a_n\right| r^{n+1}}\] |
|
