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When I'm reading Complex Analysis by Elias M.Stein, I met a question on the proof of the fundamental theorem of algebra. Stein says,

Since each term in the parentheses goes to $0$ as $\vert z\vert \to \infty$ we conclude that there exists $R>0 $ so that if $c=\vert a_n\vert /2$, then \begin{array}{cc} | P(z)|\geq c| z|^n, \text{ whenever } |z|>R.\\ \end{array} where $P(z)$ is a non-constant polynomial $P(z)=a_nz^n+\cdots+a_0$ with complex coefficients.

I searched on wiki and found what he said is likely called growth lemma, I could hardly find something helpful about "growth lemma". I'm curious about its proof. I don't know how could they, Stein and wiki, come to this conclusion. Why $c$ is $|a_n|/2$ but not a quarter or else? Who could help me?

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  • $\begingroup$ It doesn't matter whether $c = |a_n|/2$ or $c = |a_n|/4$ or $c = 0.9999\, |a_n|$; all those versions are true (with different choices of $R$). It does matter that $c > 0$ and $c < |a_n|$, as otherwise the lemma is useless, or false (respectively). $\endgroup$ – Erick Wong Jan 25 '14 at 15:41
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His statement is a precise way of saying that the leading term of a polynomial dominates for large enough $|z|$. Assume $a_{n}\ne 0$. For the polynomial in question, whenever $0 < r \le |z|$, one has $1 \le |z|/r$ and $$ \begin{align} |a_{n-1}z^{n-1}+\cdots+a_{0}| & \le |a_{n-1}||z|^{n-1}+\cdots+|a_{0}| \\ & \le \left(\frac{|a_{n-1}|}{r} +\frac{|a_{n-2}|}{r^{2}} +\cdots+\frac{|a_{0}|}{r^{n}}\right)|z|^{n} \end{align} $$ The term in parentheses in the last expression tends to 0 as $r\rightarrow \infty$. So you can choose $R$ large enough that this term is bounded by $|a_{n}|/2$ for $r > R$. Then, for $r > R$, $$ |P(z)| \ge |a_{n}||z|^{n}-|a_{n-1}z^{n-1}+\cdots a_{0}| > |a_{n}||z|^{n}-\frac{|a_{n}|}{2}|z|^{n} \ge \frac{|a_{n}|}{2}|z|^{n},\;\; |z| > R. $$

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Growth lemmas say that some "complicated" functions grows (when the variable tends to...) like some "simple" function.

In this case, the dominant (bigger) term of $P(z)$ when $z\to\infty$ is $a_n z^n$, that is, $P(z)=a_n z^n+$"small garbage" and $|P(z)|\ge|a_n||z|^n - |{\rm "small~garbage"}|\ge c|a_n||z|^n$ for any $c$ s.t. $0<c<1$ and $|z|$ large enough.

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