shows that if $P(z)$ has no complex zero, then $\frac{1}{P(z)}$ is bounded

Question

I hope someone could enlighten me!

If $P(z)$ is a polynomial, shows that if $P(z)$ has no complex zero, then $\frac{1}{P(z)}$ is bounded.

Answer 1

Amitesh Datta's outlined proof is absolutely correct. Let me give another approach, which is in some sense more direct (e.g. it is not a proof by contradiction). The differences between his answer (given first!) and mine are relatively minor.

We may assume that $P(z)$ is nonconstant.

Step 1: For any nonconstant polynomial function $P(z)$, $\lim_{z \rightarrow \infty} |P(z)| = \infty$. (This easily reduces to the case of $z$ a real variable, in which case it is familiar from calculus.)

Step 2: If for all $z \in \mathbb{C}$ $P(z) \neq 0$, then the function $f(z) = \frac{1}{P(z)}$ is a continuous [indeed holomorphic, but we won't need this] function from $\mathbb{C}$ to $\mathbb{C}$. Moreover, by Step 1, $\lim_{z \rightarrow \infty} f(z) = 0$.

Step 3: I claim that any continuous function $f: \mathbb{C} \rightarrow \mathbb{C}$ which vanishes at infinity -- i.e., has $\lim_{z \rightarrow \infty} f(z) = 0$ -- is bounded. Indeed, there exists $R_1 > 0$ such that for all $z$ with $|z| \geq R_1$, $|f(z)| \leq 1$. Moreover, since $\{z \in \mathbb{C} \ | \ |z| \leq R_1 \}$ is a closed disk in the complex plane, it is compact and thus every continuous function on $R_1$ is bounded: there exists $M \in \mathbb{R}$ such that for all $z$ with $|z| \leq R_1$, $|f(z)| \leq M$. It follows that $f$ is bounded on all of $\mathbb{C}$ by $\max \{M,1\}$.

Here is another way to construe Step 3: for any locally compact space $X$, one can give a meaning to a function $f: X \rightarrow \mathbb{C}$ vanishing at infinity: it means that for every $\epsilon > 0$, there exists a compact subset $K \subset X$ such that $|f(x)| \leq \epsilon$ for all $x \in X \setminus K$. But in fact this is equivalent to saying that $f$ extends continuously to the one-point compactification $X_{\infty} = X \cup \{\infty \}$ of $X$ with $f(\infty) = 0$. Then, since $f$ extends continuously to a function with a compact domain, it is bounded. Note that in this case the one-point compactification of $\mathbb{C}$ is nothing else than the Riemann sphere, and this way of looking at complex functions plays a big role in complex analysis.

Answer 2

Let us assume, for a contradiction, that the function defined by the rule $f(z)=\frac{1}{p(z)}$ is not bounded. In this case, there exists a complex number $z_n$ such that $\left|f(z_n)\right|>n$ for all positive integers $n$. Of course, this implies that $\left|p(z_n)\right|<\frac{1}{n}$ for all positive integers $n$.

Exercise 1: Prove that $\lim_{\left|z\right|\to\infty} \left|p(z)\right|=\infty$.

Exercise 2: Prove that the sequence $\{z_n\}_{n\in\mathbb{N}}$ is bounded. (Hint: use Exercise 1.)

Exercise 3: Prove that the sequence $\{z_n\}_{n\in\mathbb{N}}$ has a convergent subsequence. (Hint: use Exercise 2.)

Exercise 4: Prove that $z\to p(z)$ is a continuous function from $\mathbb{C}\to\mathbb{C}$.

Exercise 5: Deduce that $p$ has a complex zero and hence obtain a contradiction. (Hint: use Exercise 3 and Exercise 4.)

I hope this helps!

Answer 3

"if $p(z)$ has no complex zero, then $\frac{1}{p(z)}$ is bounded"

Not really; the exponential function is one of the simplest counterexamples...