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

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.

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You mean if $P(z)$ is a polynomial. – André Nicolas Jul 16 '11 at 2:46
Is there a difference between $P(z)$ and $p(z)$? Are you assuming your function is continuous? And, if this is homework, what have you tried so far? – Amit Kumar Gupta Jul 16 '11 at 2:50
sorry,i think this was in the lemma that trying to prove fundamental theorem of algebra – Victor Jul 16 '11 at 2:53
sorry, how can I add a comment, I am new here. Are complex zero and real zero same, both 0i+0=0; – Heather Jul 16 '11 at 2:58
@Victor I have answered your question in the form of a series of Exercises. – Amitesh Datta Jul 16 '11 at 3:18

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!

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Excuse me,i can't understand how you finished exercise 4 and 5 and how it help? – Victor Jul 16 '11 at 3:15
We assume, for a contradiction, that $p$ is not bounded and in Exercise 5 we deduce that this leads to the result that $p$ has a complex zero. (Let us recall that this is a contradiction since we explictly assumed in the beginning that $p$ has no complex zeros.) Please do think about Exercise 4 and Exercise 5 since they are not too difficult to solve. – Amitesh Datta Jul 16 '11 at 3:26
@Victor Hint (Exercise 4): prove that if $h$ and $k$ are continuous functions from $\mathbb{C}\to\mathbb{C}$, then so is $ah+bk$ for all $a,b\in\mathbb{C}$. Also, prove that $z\to z^n$ is a continuous function from $\mathbb{C}\to\mathbb{C}$ for all nonnegative integers $n$. – Amitesh Datta Jul 16 '11 at 9:23
@Theo I absolutely agree with you but perhaps the OP is doing a university course in complex analysis and has already completed a course in real analysis (?). In this case, there might be no way (short of dropping the course) for the OP to learn real analysis "properly" before learning complex analysis. – Amitesh Datta Jul 16 '11 at 9:52

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.

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+1 A very nice answer (as usual) since it introduces the OP to the Riemann sphere which is important. I think that this argument is more straightforward than mine because mine secretly uses sequential compactness (which might not be familiar to the OP). – Amitesh Datta Jul 16 '11 at 7:35
@Pete: In step 1, I think you meant $P(z)$ should be a non -constant polynomial function. – Zev Chonoles Jul 16 '11 at 8:23
In step 1 you mean non-constant. Also, I prefer to prove this via the growth lemma for polynomials, which is just the triangle inequality instead of using calculus. (Ah, I see @Zev beat me to it, but I leave my comment because of the alternative argument). – t.b. Jul 16 '11 at 8:26
@Theo: well, technically speaking I didn't say that one should use calculus to prove it but only that the real variable case is familiar from calculus. In all honesty, looking deep into my heart, whatever way the reader wants to prove this fact is absolutely fine with me. :) – Pete L. Clark Jul 16 '11 at 8:37
:) +1 ${}{}{}{}$ – t.b. Jul 16 '11 at 8:40

"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...

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