# Proofs of the Fundamental Theorem of Algebra

## Argument Principle

• 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 $$N(0) = n$$.

## Rouche’s Theorem

• 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.

## Liouville’s Theorem

• 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. $$\contradiction$$

## Open Mapping Theorem

• $$p$$ 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.

## Generalized Liouville

If $$X$$ is a compact complex manifold, any holomorphic $$f:X\to {\mathbf{C}}$$ is constant.

If $$f:X\to Y$$ is a nonconstant holomorphic map between Riemann surfaces with $$X$$ compact, then

• $$f$$ must be surjective,
• $$Y$$ must be compact,
• $$f^{-1}(q)$$ is finite for all $$q\in Y$$,
• The branch and ramification loci consist of finitely many points.

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 $$p(x) = 0$$.