Proof of the Fundamental Theorem of Algebra: filling in some intermediate steps

complex-analysispolynomialsrouches-theorem

I'm familiar with Rouche's theorem in the following form:

If $f, g$ are analytic on a domain $\Omega$ with $|g(z)| < |f(z)|$ on $\partial \Omega$, then $f$ and $f+g$ have the same number of zeros in $\Omega$.

I'm walking through how to use this to prove the fundamental theorem of algebra, but am stuck on how the "usual" lower bound is used and where it comes from. The argument goes something like this:

  1. Suppose $p(z) = a_n z^n + \cdots + a_1 z + a_0$
  2. Set $f(z) = a_n z^n$
  3. Set $g(z) = a_{n-1}z^{n-1} + \cdots + a_1 z + a_0$
  4. Choose $R$ large enough such that

$$R > \max\left( {|a_{n-1}| + \cdots + |a_1| + |a_0| \over |a_n| }, 1\right)$$

  1. Then $|g(z)| < |f(z)|$ on the circle $|{z}|= R$, so apply Rouche and note that $z^n$ has $n$ zeros at $z_0 = 0$, which is in this region.

I am stuck on part 4: I buy that such an $R$ can be chosen, since these coefficients are fixed. What I can't work out is the explicit inequality that shows 4 $\implies$ 5.

The argument I'm looking for would essentially fill in the vdots in this chain of inequalities:

$$\begin{align*}
|g(z)|
&:= |a_{n-1}z^{n-1} + \cdots + a_1 z + a_0 | \\
&\leq |a_{n-1}z^{n-1}| + \cdots + |a_1 z| + |a_0 | \quad\text{by the triangle inequality} \\
&= |a_{n-1}|\cdot |z^{n-1}| + \cdots + |a_1|\cdot| z| + |a_0 | \\
& \quad\vdots \quad ? \\
&\leq |a_n z^n| \\
&:= |f(z)|
\end{align*}$$

I would prefer a proof that doesn't necessarily assume that $p$ is monic, since I'd like to trace $a_n$ throughout the inequality.

I would also welcome answers that show how you might run this problem backwards, i.e. writing out some inequalities in order to deduce what $R$ should be.

Best Answer

$$\begin{align*} |a_{n-1}|\cdot |z^{n-1}| + \cdots + |a_1|\cdot| z| + |a_0 | &= |a_{n-1}|\cdot R^{n-1} + \cdots + |a_1| R + |a_0 | \\ &\leq |a_{n-1}|\cdot R^{n-1}+|a_{n-2}|\cdot R^{n-1} + \cdots + |a_1| \cdot R^{n-1} + |a_0 |\cdot R^{n-1} \qquad\text{(1)}\\ &= R^{n-1} \left( |a_{n-1}| + |a_{n-2}| + \cdots + |a_1| + |a_0| \right) \\ &\leq R^{n-1} \left( |a_n|\cdot R \right) \qquad\text{(2)}\\ &= R^{n} |a_n| \end{align*} $$

In inequality (1) I have used the fact that $R^{i} \leq R^{n-1}$ for $0 \leq i \leq n-1$ and inequality (2) follows by choice of $R$.

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