Fundamental theorem of algebra

Today we'll prove the fundamental theorem of algebra: that every nonconstant polynomial (in one variable) has a root (in $\mathbb C$).

The main result we'll be using today is Liouville's theorem, stated below.

Result 1. A bounded entire function is constant.

We won't prove Liouville's theorem, but we will prove the following results.

Result 2. The function $\mathbb C \rightarrow \mathbb C$ given by a polynomial in $z$ is entire. (A function is said to be entire if it has domain $\mathbb C$ and is holomorphic in all of $\mathbb C$.)
Result 3. If $f : U \rightarrow\mathbb C$ is holomorphic on an open subset $U$ of $\mathbb C$ and has no zeroes, then the function $g$ such that $g(z) = 1/f(z)$ is holomorphic on $U$.
Result 4. As $|z| \rightarrow \infty$, we have $|p(z)| \rightarrow \infty$.

Before we prove these, though, we will present our proof of the fundamental theorem of algebra, as given on page 87 in the complex analysis book by Greene and Krantz.

Suppose a nonconstant polynomial $p$ did not have a root in $\mathbb C$. Then the function $g$ where $g(z) = 1/p(z)$ is entire (by results 2 and 3). By result 4, we know $|p(z)| \rightarrow \infty$ as $|z| \rightarrow \infty$, meaning $|g(z)| \rightarrow 0$ as $|z| \rightarrow \infty$. This means $g$ is bounded and entire, so by Liouville's theorem, $g$ is constant, a contradiction.

With our cute proof out of the way, we'll now focus on some of our messier results.

Proof of result 2: What we have to prove here depends on our initial definition of holomorphic. We'll assume the result (or definition) that says it suffices to prove that the polynomial function is complex differentiable. By the linearity of differentiation, it suffices to show that each individual $z \mapsto z^k$ for integers $k \geq 0$ is complex differentiable.

Now, the identity map is continuous, so $\lim_{z \to z_0} z = z_0$. Using this and the properties of limits gives us
$$\lim_{z \to z_0} z^{k - 1} + z^{k - 2} z_0 + ... + z_0^{k - 1} = kz_0^{k - 1}.$$Then we see that
$$\displaystyle\lim_{z \rightarrow z_0} \frac{z^k - z_0^k}{z - z_0}$$exists for all $k \geq 0$, since
$$(z^k - z_0^k) / (z - z_0) = z^{k - 1} + z^{k - 2} z_0 + ... + z_0^{k - 1}.$$

Proof of result 3: Let $z_0 \in U$. Then
\[
\begin{align*}
 \displaystyle\lim_{z \rightarrow z_0} \frac{ \frac1{f(z)} - \frac1{f(z_0)} }{z - z_0} &= \lim_{z \rightarrow z_0} \frac{f(z_0) - f(z)}{z - z_0} \cdot \frac 1{f(z)f(z_0)}\\
 &= \frac{-1}{f(z_0)^2} \cdot f'(z_0).
\end{align*}\]Since $g$ has a complex derivative at $z_0$ for any $z_0 \in U$, $g$ is holomorphic.

Proof of result 4: Here we write $p(z)$ as
$$z^n ( a_n + \frac {a_{n - 1}}z + ... + \frac{a_0}{z^n} ).$$ By repeatedly applying the triangle inequality,
\[
\begin{align*}
 |a_n + \frac{a_{n - 1}}{z} + ... + \frac{a_0}{z^n}|
 &\geq |a_0| - \left|\frac{a_{n - 1}}z\right| - ... - \left|\frac{a_0}{z^n}\right|\\
 &= |a_0| - \left( \frac{|a_{n - 1}|}{|z|}+ ... + \frac{|a_0|}{|z|^n} \right).
\end{align*}\]As $|z| \to \infty$, the right side of the above inequality approaches $|a_0|$. This means that we can pick an $\epsilon > 0$ such that $\epsilon < |a_0|$, and then there exists a $K$ such that $|z| > K$ implies that the right side of the above inequality is in $(|a_0| - \epsilon, |a_0| + \epsilon)$.

So, when $|z| > K$,
$$|p(z)| \geq |z|^n |a_0 - \epsilon|,$$which becomes arbitrarily large as $|z| \to \infty$. This concludes our proofs.