Section11.5Fundamental theorem of algebra

Note: half a lecture, optional

In this section we study the local behavior of polynomials and the growth of polynomials as $$z$$ goes to infinity. As an application we prove the fundamental theorem of algebra: Any nonconstant polynomial has a complex root.

Proof.

Without loss of generality assume that $$z_0 = 0$$ and $$p(0) = 1\text{.}$$ Write

\begin{equation*} p(z) = 1+a_kz^k + a_{k+1}z^{k+1} + \cdots + a_d z^d , \end{equation*}

where $$a_k \not= 0\text{.}$$ Pick $$t$$ such that $$a_k e^{ikt} = -\sabs{a_k}\text{,}$$ which we can do by the discussion on trigonometric functions. Suppose $$r > 0$$ is small enough such that $$1-r^k \sabs{a_k} > 0\text{.}$$ We have

\begin{equation*} p(r e^{it}) = 1-r^k \sabs{a_k} + r^{k+1}a_{k+1}e^{i(k+1)t} + \cdots + r^{d}a_{d}e^{idt} . \end{equation*}

So

\begin{equation*} \begin{split} \abs{ p(r e^{it}) } - \abs{ r^{k+1}a_{k+1}e^{i(k+1)t} + \cdots + r^{d}a_{d}e^{idt} } & \leq \abs{ p(r e^{it}) - r^{k+1}a_{k+1}e^{i(k+1)t} - \cdots - r^{d}a_{d}e^{idt} } \\ & = \abs{ 1-r^k \sabs{a_k} } = 1-r^k \sabs{a_k} . \end{split} \end{equation*}

In other words,

\begin{equation*} \abs{ p(r e^{it}) } \leq 1-r^k \left( \sabs{a_k} - r \abs{ a_{k+1}e^{i(k+1)t} + \cdots + r^{d-k-1}a_{d}e^{idt} } \right) . \end{equation*}

For small enough $$r\text{,}$$ the expression in the parentheses is positive as $$\sabs{a_k} > 0\text{.}$$ Hence, $$\abs{p(re^{it})} < 1 = p(0)\text{.}$$

Remark11.5.2.

The lemma above holds essentially with an unchanged proof for (complex) analytic functions. A proof of this generalization is left as an exercise to the reader. What the lemma says is that the only minima the modulus of analytic functions (polynomials) has are precisely at the zeros.

Remark11.5.3.

The lemma does not hold if we restrict to real numbers. For example, $$x^2+1$$ has a minimum at $$x=0\text{,}$$ but no zero there. The thing is that there is a $$w$$ arbitrarily close to $$0$$ such that $$\sabs{w^2+1} < 1\text{,}$$ but this $$w$$ is necessarily not real. Letting $$w = i\epsilon$$ for small $$\epsilon > 0$$ works.

The moral of the story is that if $$p(0) = 1\text{,}$$ then very close to 0, the polynomial looks like $$1+az^k\text{,}$$ and $$1+az^k$$ has no minimum at the origin. All the higher powers of $$z$$ are too small to make a difference. We find similar behavior at infinity.

Proof.

Write $$p(z) = a_0 + a_1 z + \cdots + a_d z^d$$ and suppose that $$d \geq 1$$ and $$a_d \not= 0\text{.}$$ Suppose $$\sabs{z} \geq R$$ (so also $$\sabs{z}^{-1} \leq R^{-1}$$). We estimate:

\begin{equation*} \begin{split} \sabs{p(z)} & \geq \sabs{a_d z^d} - \sabs{a_0} - \sabs{a_1 z} - \cdots - \sabs{a_{d-1} z^{d-1} } \\ & = \sabs{z}^d \bigl( \sabs{a_d} - \sabs{a_0} \, \sabs{z}^{-d} - \sabs{a_1} \, \sabs{z}^{-d+1} - \cdots - \sabs{a_{d-1}} \, \sabs{z}^{-1} \bigr) \\ & \geq R^d \bigl(\sabs{a_d} - \sabs{a_0}R^{-d} - \sabs{a_1}R^{1-d} - \cdots - \sabs{a_{d-1}}R^{-1} \bigr) . \end{split} \end{equation*}

Then the expression in parentheses is eventually positive for large enough $$R\text{.}$$ In particular, for large enough $$R$$ we get that this expression is greater than $$\frac{\sabs{a_d}}{2}\text{,}$$ and so

\begin{equation*} \sabs{p(z)} \geq R^d \frac{\sabs{a_d}}{2} . \end{equation*}

Therefore, we can pick $$R$$ large enough to be bigger than a given $$M\text{.}$$

The lemma above does not generalize to analytic functions, even those defined in all of $$\C\text{.}$$ The function $$\cos(z)$$ is a counterexample. Note that we had to look at the term with the largest degree, and we only have such a term for a polynomial. In fact, something that we will not prove is that an analytic function defined on all of $$\C$$ satisfying the conclusion of the lemma must be a polynomial.

The moral of the story here is that for very large $$\sabs{z}$$ (far away from the origin) a polynomial of degree $$d$$ really looks like a constant multiple of $$z^d\text{.}$$

Proof.

Let $$\mu := \inf \bigl\{ \sabs{p(z)} : z \in \C \bigr\}\text{.}$$ Find an $$R$$ such that for all $$z$$ with $$\sabs{z} \geq R\text{,}$$ we have $$\sabs{p(z)} \geq \mu+1\text{.}$$ Therefore, every $$z$$ with $$\sabs{p(z)}$$ close to $$\mu$$ must be in the closed ball $$C(0,R) = \bigl\{ z \in \C : \sabs{z} \leq R \bigr\}\text{.}$$ As $$\sabs{p(z)}$$ is a continuous real-valued function, it achieves its minimum on the compact set $$C(0,R)$$ (closed and bounded) and this minimum must be $$\mu\text{.}$$ So there is a $$z_0 \in C(0,R)$$ such that $$\sabs{p(z_0)} = \mu\text{.}$$ As that is a minimum of $$\sabs{p(z)}$$ on $$\C\text{,}$$ then by the first lemma above, we have $$\sabs{p(z_0)} = 0\text{.}$$

The fundamental theorem also does not generalize to analytic functions. For example, $$e^{z}$$ is an analytic function on $$\C$$ with no zeros.

Subsection11.5.1Exercises

Exercise11.5.1.

Prove Lemma 11.5.1 for an analytic function. That is, suppose that $$p(z)$$ is a power series around $$z_0\text{.}$$

Exercise11.5.2.

Use Exercise 11.5.1 to prove the maximum principle for analytic functions: If $$U \subset \C$$ is open and connected, $$f \colon U \to \C$$ is analytic, and $$\sabs{f(z)}$$ attains a relative maximum at $$z_0 \in U\text{,}$$ then $$f$$ is constant.

Exercise11.5.3.

Let $$U \subset \C$$ be open and $$z_0 \in U\text{.}$$ Suppose $$f \colon U \to \C$$ is analytic and $$f(z_0) = 0\text{.}$$ Show that there exists an $$\epsilon > 0$$ such that either $$f(z) \not= 0$$ for all $$z$$ with $$0 < \sabs{z} < \epsilon$$ or $$f(z) = 0$$ for all $$z \in B(z_0,\epsilon)\text{.}$$ In other words, zeros of analytic functions are isolated. Of course, same holds for polynomials.

A rational function is a function $$f(z) := \frac{p(z)}{q(z)}$$ where $$p$$ and $$q$$ are polynomials and $$q$$ is not identically zero. A point $$z_0 \in \C$$ where $$f(z_0) = 0$$ (and therefore $$p(z_0) = 0$$) is called a zero. A point $$z_0 \in \C$$ is called an singularity of $$f$$ if $$q(z_0) = 0\text{.}$$ As all zeros are isolated and so all singularities of rational functions are isolated and so are called an isolated singularity. An isolated singularity is called removable if $$\lim_{z \mapsto z_0} f(z)$$ exists. An isolated singularity is called a pole if $$\lim_{z \mapsto z_0} \sabs{f(z)} = \infty\text{.}$$ We say $$f$$ has pole at $$\infty$$ if

\begin{equation*} \lim_{z \to \infty} \sabs{f(z)} = \infty , \end{equation*}

that is, if for every $$M > 0$$ there exists an $$R > 0$$ such that $$\sabs{f(z)} > M$$ for all $$z$$ with $$\sabs{z} > R\text{.}$$

Exercise11.5.4.

Show that a rational function which is not identically zero has at most finitely many zeros and singularities. In fact, show that if $$p$$ is a polynomial of degree $$n > 0$$ it has at most $$n$$ zeros.
Hint: If $$z_0$$ is a zero of $$p\text{,}$$ without loss of generality assume $$z_0 = 0\text{.}$$ Then use induction.

Exercise11.5.5.

Prove that if $$z_0$$ is a removable singularity of a rational function $$f(z) := \frac{p(z)}{q(z)}\text{,}$$ then there exist polynomials $$\widetilde{p}$$ and $$\widetilde{q}$$ such that $$\widetilde{q}(z_0) \not= 0$$ and $$f(z) = \frac{\widetilde{p}(z)}{\widetilde{q}(z)}\text{.}$$
Hint: Without loss of generality assume $$z_0 = 0\text{.}$$

Exercise11.5.6.

Given a rational function $$f$$ with an isolated singularity at $$z_0\text{,}$$ show that $$z_0$$ is either removable or a pole.
Hint: See the previous exercise.

Exercise11.5.7.

Let $$f$$ be a rational function and $$S \subset \C$$ is the set of the singularities of $$f\text{.}$$ Prove that $$f$$ is equal to a polynomial on $$\C \setminus S$$ if and only if $$f$$ has a pole at infinity and all the singularities are removable.
Hint: See previous exercises.

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