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I'm learning about complex analysis and need some help with this problem:

If $f: \mathbb{C} \rightarrow \mathbb{C}$ is analytic and $\lim_{z \to \infty} f(z) = \infty$ show that $f$ is a polynomial (hint: consider the function $g(z) = f(1/z)$).

Recall that poles are points where evaluating the function would entail dividing by zero. Therefore, since $\lim_{z \to \infty} f(z) = \infty$ this means that $\infty$ is a pole of $f$. How do I continue from here and make use of the hint?


I should mention that this problem has already been asked by other members but I could not find any solution using the given hint.

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  • $\begingroup$ @Christopher said everything. Look what happens when $z \to \infty$ on series. $\endgroup$ – L.F. Cavenaghi Feb 9 '16 at 23:37
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Suppose $f$ has Taylor series $$ f(z)=\sum_{n=0}^{\infty}a_nz^n \quad\text{and }\quad g(z) =f(1/z) =\sum_{n=0}^{\infty}\frac{a_n}{z^n} $$ If $f$ is not polynomial, then $0$ is an essential singularity of $g$ ($\infty$ is an essential singularity of $f$). By Casorati-Weierstrass theorem, for any $A\in \Bbb{C}$, there is a sequence $z_n\to0$ such that $\lim_{n\to\infty}g(z_n)=A$, i.e. there is $z_n'=1/z_n\to\infty$ such that $\lim_{n\to\infty}f(z_n')=A$, contradicting $\lim_{n\to\infty}f(z)=\infty$.

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Hint: Expand $g(z)$ by Laurent series.

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    $\begingroup$ Does that help without using Picard's theorem on essential singularities? Or is there some simple way of seeing that $g(z)$ has a removable singularity at $0$? $\endgroup$ – Rob Arthan Feb 10 '16 at 0:10
  • $\begingroup$ @RobArthan $g$ doesn't have a removable singularity, it has a pole, but yes, that requires some knowledge about why it can't be an essential singularity (Casorati-Weierstrass would suffice). But based on the original question it sounds like the OP was aware of that conclusion. $\endgroup$ – Christopher A. Wong Feb 10 '16 at 0:17
  • $\begingroup$ Sorry, that was a sort of semantic typo: I meant "pole" not "removable singularity". Yes, C-W is enough, but is easily forgotten (at least by me) once you know Picard's theorem. Let's leave it to the OP to see what they can make of your hint. $\endgroup$ – Rob Arthan Feb 10 '16 at 0:30
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Without Laurent series and assuming $\;f(z)\;$ isn't zero (because then it is trivially true).

By the given information there exists $\;M\in\Bbb R^+\;$ such that $\;|f(z)|>1\;\;\;\forall\,z\in\Bbb C\;\;\text{with}\;\;|z|>M\;$ .

It must be that $\;f(z)\;$ has a finite number of zeros $\;z_1,...,z_n\;$, otherwise its set of zeros, which is in $\;C_M:=\{\,z\in\Bbb C\;;\;|z|\le M\}\;$, has an accumulation point by Bolzano-Weierstrass, and thus from the identity theorem this would mean $\;f(z)=0\;$ .

From here that $\;g(z):=\frac{f(z)}{\prod\limits_{k=1}^n(z-z_k)}\;$ is analytic and non-zero, and thus also $\;h(z)=\frac1{g(z)}\;$ is, and we have for $\;z\in\Bbb C\setminus C_M\;$:

$$|h(z)|=\frac{|z^n+A|}{|f(z)|}\le|z^n|+A\implies h(z)\;\;\text{is a polynomial without roots}\;(**)\;\implies h(z)=K$$

a constant, by the Fundamental Theorem of Algebra, and thus also is $\;g(z)\;$ :

$$\frac{f(z)}{\prod\limits_{k=1}^n(z-z_k)}=g(z)=\frac1{h(z)}=\frac1K\implies f(z)=K(z-z_1)\cdot\ldots\cdot(z-z_n)$$

and we've finished.

If you need a prove of $\;(**)\;$ ask back.

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  • $\begingroup$ Possible typo :In your def'n of $g(z)$ shouldn't each term in the denominator be $(z-z_k)^{e_k}$ where each $e_k$ is a positive integer, the degree of the zero of $f$ at $z_k$? $\endgroup$ – DanielWainfleet Feb 10 '16 at 1:43
  • $\begingroup$ @user254665 Thanks you, but I don't think so: I didn't write the zeros are different, and we don't need that here. $\endgroup$ – DonAntonio Feb 10 '16 at 1:48
  • $\begingroup$ OK. I was thinking of the zeroes as the set of points where f=0. Too much set theory in my head. $\endgroup$ – DanielWainfleet Feb 10 '16 at 4:28
  • $\begingroup$ @Joanpemo How did you get $|h(z)|=\frac{|z^n+A|}{|f(z)|}$? What is $A$? $\endgroup$ – Sarah May 3 '17 at 3:05

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