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.

  • $\begingroup$ @Christopher said everything. Look what happens when $z \to \infty$ on series. $\endgroup$ – L.F. Cavenaghi Feb 9 '16 at 23:37

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 $\lim_{n\to\infty}f(z)=\infty$.


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

  • 1
    $\begingroup$ 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

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.

  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.