FTOA: Let $f(z) = |a_nz^n + .. + a_1z + a_0|$. Show $f(z)$ has a minimizer $z^*$ and complex sum converges for $|z| \rightarrow \infty$ I've been looking into this proof of the Fundamental Theorem of Algebra:
http://cuhkmath.wordpress.com/2011/06/28/another-proof-of-the-fundamental-theorem-of-algebra/
In the proof we have $f(z) = |p(z)| = |a_nz^n + ... + a_1z + a_0|$ and we show $f(z) \rightarrow \infty$ when $|z| \rightarrow \infty$ by claiming $$|a_n + \frac {a_{n-1}} z + ... + \frac {a_1} {z^{n-1}} + \frac {a_0} {z^{n}}|$$ converges to $a_n$ when $|z| \rightarrow \infty$
This in turn implies $f(z)$ has a minimizer $z^* \in \mathbb C$.
Can someone show formally why the sum in modulus converges to $a_n$? And why this implies $f(z)$ has a minimizer $z^* \in \mathbb C$ ?
 A: By the triangle inequality, we have
$$\left| \left(a_n + \frac{a_{n-1}}{z} + \ldots + \frac{a_0}{z^n}\right) - a_n \right| \leq \left| \frac{a_{n-1}}{z} \right| + \ldots+ \left| \frac{a_0}{z^n} \right|. \tag{1}$$
Let $a := \max\{|a_0|,\ldots,|a_{n-1}|\}$, then $(1)$ shows
$$\left| \left(a_n + \frac{a_{n-1}}{z} + \ldots + \frac{a_0}{z^n}\right) - a_n \right| \leq  \frac{n \cdot a}{|z|}$$
for any $|z| \geq 1$. Letting $|z| \to \infty$ we find
$$\left| \left(a_n + \frac{a_{n-1}}{z} + \ldots + \frac{a_0}{z^n}\right) - a_n \right| \to 0$$
i.e.
$$\left(a_n + \frac{a_{n-1}}{z} + \ldots + \frac{a_0}{z^n}\right) \to a_n. \tag{2}$$
This proves the first claim. By definition,
$$f(z) = \left|z^n \cdot \left(a_n + \frac{a_{n-1}}{z} + \ldots + \frac{a_0}{z^n}\right) \right|$$
and therefore $(2)$ implies $f(z) \to \infty$ as $|z| \to \infty$. In particular, we can choose an arbitrary $z_0 \in \mathbb{C}$ and find $R \geq 0$ such that $f(z) \geq f(z_0)$ for any $|z| \geq R$. Since $f$ is continous, hence attains its minimum on compact sets, there exists $z^\ast \in \mathbb{C}$, $|z^\ast| \leq R$, such that $$f(z^{\ast}) = \min_{z \in \mathbb{C}} f(z)$$
