2
$\begingroup$

Let $f(z)$ be an analytic function in the whole complex plane apart from simple poles at $z_1,...,z_m$. Moreover $f(1/z)$ has a simple pole at $0$. Show that $f$ is a ratio of two polynomials.

To start this, we know $f(z)=\frac{A_1}{z-z_1}+...+\frac{A_m}{z-z_m}+h(z)$ where $h(z)$ is in the form $\sum_{k=1}^\infty a_nz^n$. Since $f(1/z)$ has a simple pole at zero. Then $h(z)$ must be polynomial. So $f(z)$ is a ratio of polynomials. Is this true and clear? Thanks

$\endgroup$
2
$\begingroup$

Yes, it is true that if $h(1/z)$ has a simple pole at $z=0$, then $h(z)$ must be a polynomial but why? I think you need more argumentation. I like better to proceed in a slightly different way:

Consider the function $F: \mathbb{C} \to \mathbb{C}$ given by $$ F(z) = f(z) \prod_{j=1}^m (z-z_j) $$ Clearly $F$ is an entire function, since we are "removing" the poles of $f$. Since $f(1/s)$ has a simple pole at $s=0$, then also $F(1/s)$ has it, therefore it's Laurent series in the annulus $\{ 0< |s|<r \}$ for some $r>0$ is $$ F(1/s) = \sum_{n\boldsymbol{= -}1}^{\infty} a_n s^n $$ Hence $s\mapsto sF(1/s)$ is analytic in $\{ 0< |s|<r \}$, thus there exist $M>0$ such that $|sF(1/s)|\leq M$ if $s \in \{ 0< |s|<r \}$. Putting $z=1/s$ and $R=1/r$, we get $$ |F(z) | \leq M |z|\ \text{ for $|z|>R$} \tag{1} $$ But since $F$ is entire, line $(1)$ together with Cauchy estimates gives that $F$ must be a polynomial with degree at most 1. Therefore, indeed $f$ is a ratio of polynomials since $$ f(z) = \frac{F(z)}{ \prod_{j=1}^m (z-z_j)} $$

$\endgroup$
  • 1
    $\begingroup$ Thanks so much. Thats a much better argument $\endgroup$ – nerd Jul 14 '15 at 16:28
  • $\begingroup$ @nerd You are welcome and thanks !! $\endgroup$ – Alonso Delfín Jul 14 '15 at 16:39
  • 1
    $\begingroup$ @nerd if this answers your question satisfactorily remember to accept the answer. $\endgroup$ – hjhjhj57 Jul 14 '15 at 17:33

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.