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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.2011-07-16
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    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?2011-07-16
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    sorry,i think this was in the lemma that trying to prove fundamental theorem of algebra2011-07-16
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    @Victor I have answered your question in the form of a series of ***Exercises***.2011-07-16
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    sorry, how can I add a comment, I am new here. Are complex zero and real zero same, both 0i+0=0;2011-07-16
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    @onlybluemoon: 1) you can't comment on anything other than your own questions until you reach a certain reputation level. This is to prevent excessive spam. 2) if you want to ask a new question, ask it in a separate question.2011-07-16

3 Answers 3

2

"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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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).2011-07-16
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    @Pete: In step 1, I think you meant $P(z)$ should be a *non* -constant polynomial function.2011-07-16
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    In step 1 you mean non-constant. Also, I prefer to prove this via the [growth lemma](http://math.stackexchange.com/questions/39567/for-a-polynomial-pz-prove-there-exist-r0-such-that-if-z-r-then-p/39579#39579) 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).2011-07-16
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    @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. :)2011-07-16
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    :) +1 ${}{}{}{}$2011-07-16
4

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?2011-07-16
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    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.2011-07-16
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    @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 $n$.2011-07-16
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    @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.2011-07-16