Proofs of the Fundamental Theorem of Algebra

• Let $P(z) = a_nz^n + \cdots + a_0$ and $g(z) = P'(z)/P(z)$, note $P$ is holomorphic
• Since $\lim_{{\left\lvert {z} \right\rvert} \to \infty} P(z) = \infty$, there exist an $R>0$ such that $P$ has no roots in $\left\{{{\left\lvert {z} \right\rvert} \geq R}\right\}$.
• Apply the argument principle:

  \begin{align*}
  N(0) = {1\over 2\pi i} \oint_{{\left\lvert {\xi} \right\rvert} = R} g(\xi) \,d\xi
  .\end{align*}

• Check that $\lim_{{\left\lvert {z\to \infty} \right\rvert}}zg(z) = n$, so $g$ has a simple pole at $\infty$
• Then $g$ has a Laurent series ${n\over z} + {c_2 \over z^2} + \cdots$
• Integrate term-by-term to get $N(0) = n$.

• Let $P(z) = a_nz^n + \cdots + a_0$
• Set $f(z) = a_n z^n$ and $g(z) = P(z) - f(z) = a_{n-1}z^{n-1} + \cdots + a_0$, so $f+g = P$.
• Choose $R > \max\qty{ { {\left\lvert {a_{n-1}} \right\rvert} + \cdots + {\left\lvert {a_0} \right\rvert} \over {\left\lvert {a_n} \right\rvert} }, 1}$, then

\begin{align*}
|g(z)| 
&\coloneqq|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 | \\
&=  |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} \quad\text{since } R>1 \implies R^{a+b} \geq R^a \\
&= 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) \quad\text{by choice of } R   \\
&= R^{n} |a_n| \\
&= |a_n z^n| \\
&\coloneqq|f(z)|
\end{align*}

• Then $a_n z^n$ has $n$ zeros in ${\left\lvert {z} \right\rvert} < R$, so $f+g$ also has $n$ zeros.

• Suppose $p$ is nonconstant and has no roots, then ${1\over p}$ is entire. We will show it is also bounded and thus constant, a contradiction.
• Write $p(z) = z^n \left(a_n + \frac{a_{n-1}}{z}+\dots+\frac{a_{0}}{z^{n}}\right)$
• Outside a disc:
  • Note that $p(z) \overset{z\to \infty }\to \infty$. so there exists an $R$ large enough such that ${\left\lvert {p(z)} \right\rvert} \geq {1\over A}$ for any fixed chosen constant $A$.
  • Then ${\left\lvert { 1/p(z)} \right\rvert} \leq A$ outside of ${\left\lvert {z} \right\rvert} >R$, i.e. $1/p(z)$ is bounded there.
• Inside a disc:
  • $p$ is continuous with no roots and thus must be bounded below on ${\left\lvert {z} \right\rvert} < R$.
  • $p$ is entire and thus continuous, and since $\overline{D}_r(0)$ is a compact set, $p$ achieves a min $A$ there
  • Set $C \coloneqq\min(A, B)$, then ${\left\lvert {p(z)} \right\rvert} \geq C$ on all of ${\mathbf{C}}$ and thus ${\left\lvert {1/p(z)} \right\rvert} \leq C$ everywhere.
  • So $1/p(z)$ is bounded an entire and thus constant by Liouville’s theorem – but this forces $p$ to be constant. $\contradiction$

• $p$ induces a continuous map ${\mathbf{CP}}^1 \to {\mathbf{CP}}^1$
• The continuous image of compact space is compact;
• Since the codomain is Hausdorff space, the image is closed.
• $p$ is holomorphic and non-constant, so by the Open Mapping Theorem, the image is open.
• Thus the image is clopen in ${\mathbf{CP}}^1$.
• The image is nonempty, since $p(1) = \sum a_i \in {\mathbf{C}}$
• ${\mathbf{CP}}^1$ is connected
• But the only nonempty clopen subset of a connected space is the entire space.
• So $p$ is surjective, and $p^{-1}(0)$ is nonempty.
• So $p$ has a root.

Given a nonconstant $p\in {\mathbf{C}}[x]$, regard it as a function $p: {\mathbf{P}}^1({\mathbf{C}}) \to {\mathbf{P}}^1({\mathbf{C}})$ by extending so that $p(\infty) = \infty$. Since $p$ is nonconstant, by the lemma $p$ is surjective, so there exists some $x\neq \infty$ in ${\mathbf{P}}^1({\mathbf{C}})$ with $p(x) = 0$.