|
|
Author |
hbghlyj
Posted at 2022-9-11 09:55:11
Theorem 7.14. Suppose that $p(z)=\sum_{k=0}^n a_k z^k$ is a non-constant polynomial where $a_k \in \mathbb{C}$ and $a_n \neq 0$. Then there is a $z_0 \in \mathbb{C}$ for which $p\left(z_0\right)=0$.
Proof. By rescaling $p$ we may assume that $a_n=1$. If $p(z) \neq 0$ for all $z \in \mathbb{C}$ it follows that $f(z)=1 / p(z)$ is an entire function (since $p$ is clearly entire). We claim that $f$ is bounded. Indeed since it is continuous it is bounded on any disc $\bar{B}(0, R)$, so it suffices to show that $|f(z)| \rightarrow 0$ as $z \rightarrow \infty$, that is, to show that $|p(z)| \rightarrow \infty$ as $z \rightarrow \infty$. But we have
\[
|p(z)|=\left|z^n+\sum_{k=0}^{n-1} a_k z^k\right|=\left|z^n\right|\left|1+\sum_{k=0}^{n-1} \frac{a_k}{z^{n-k}}\right|\geq\left|z^n\right| \cdot\left(1-\sum_{k=0}^{n-1} \frac{\left|a_k\right|}{|z|^{n-k}}\right) .
\]
Since $\frac{1}{|z|^m} \rightarrow 0$ as $|z| \rightarrow \infty$ for any $m \geq 1$ it follows that for sufficiently large $|z|$, say $|z| \geq R$, we will have $1-\sum_{k=0}^{n-1} \frac{\left|a_k\right|}{|z|^{n-k}} \geq 1 / 2$. Thus for $|z| \geq R$ we have $|p(z)| \geq \frac{1}{2}|z|^n$. Since $|z|^n$ clearly tends to infinity as $|z|$ does, it follows $|p(z)| \rightarrow \infty$ as required. $\square$
Remark 7.15. The crucial point of the above proof is that one term of the polynomial–the leading term in this case–dominates the behaviour of the polynomial for large values of $z$. All proofs of the fundamental theorem hinge on essentially this point. Note that $p(z_0)=0$ if and only if $p(z)=(z-z_0) q(z)$ for a polynomial $q(z)$, thus by induction on degree we see that the theorem implies that a polynomial over $\mathbb{C}$ factors into a product of degree one polynomials.
|
|
