1
$\begingroup$

I am facing the following question:

Let $f(z)$ be a meromorphic function on $\mathbb{C}$ with simple pole at $a_1,a_2,\dots$ such that $$0<|a_1|\leq |a_2|\leq |a_3|\leq\dots$$ and let $ A_n \equiv \underset{\,\,z=a_n}{\operatorname{Res}}f(z)$, and $C_m$ be a sequence of simple closed contours.

Prove that if the following are true:

  1. $C_m$ does not pass through any $a_n$;
  2. $\lim_{m\to\infty}r_m=\infty$, where $r_m=d(C_m,0)$;
  3. there is a constant $\beta$ such that $L_m=$Length of $C_m\leq \beta r_m$;
  4. $max_{z\in C_m}|f(z)|=\circ(r_m)$.

then $$f(z)=f(0)+\sum_{n=1}^{\infty} A_n(\frac{1}{z-a_n}+\frac{1}{a_n}).$$

I tried to show the convergence of the above summation first and then by Mittag-Leffler Theorem in $\mathbb{C}$ we can get $f(z)=\sum_{n=1}^{\infty} A_n(\frac{1}{z-a_n}+\frac{1}{a_n})+g(z)$ where $g$ is entire.

But I can't show the convergence since in these assumptions, $C_m$ may encircle no poles, so they don't give any information for convergence. I think we may assume more. For example, we assume $C_m$ contains the origin. Then how to show the convergence and if we can show the convergence, how to estimate $g$ ?

$\endgroup$
0
$\begingroup$

We want to show that $$ \tag 1 f(z) = f(0) + \sum_{n=1}^\infty \alpha_n \left( \frac{1}{z-z_n} + \frac{1}{z_n} \right), $$ where

  1. $f$ is a meromorphic function with simple poles at $z_n, \, n \in \mathbb{N}$, holomorphic at $z=0$,
  2. $\alpha_n$ is the residue of $f$ at the pole $z_n$,
  3. $f$ is uniformly bounded on every circle $\mathcal C_n$ with radius $R_n$ containing all poles $z_k$ with $k \le n$, i.e. $$ \tag 2\exists A > 0 | \,\, \forall (n \in \mathbb{N}, z \in \mathcal C_n), \,\, |f(z)| \le A $$

To do that, suppose without loss of generality that the poles $z_n$ are ordered monotonically as in $$ 0 < |z_1| \le |z_2| \le \dots, $$ and fix some $z_0 \in \mathbb{C}$ such that $f$ is holomorphic at $z_0$.

Consider the integral $$ \tag 3 I_n(z_0) \equiv \oint_{\mathcal C_n} \frac{dz}{2\pi i} \frac{f(z)}{z-z_0}, $$ which applying Residue's theorem becomes $$ \tag 4 I_n(z_0) = f(z_0) + \sum_{k=1}^n \frac{\alpha_k}{z_k-z_0}. $$ In particular for $z_0=0$ we have $$ \tag 5 I_n(0) \equiv \oint_{\mathcal C_n} \frac{dz}{2\pi i} \frac{f(z)}{z} = f(0) + \sum_{k=1}^n \frac{\alpha_k}{z_k}.$$ Taking the difference between (3) and (5) (in the integral form) we see that: $$ \tag 6 |I_n(z_0) - I_n(0) | \le \oint_{\mathcal C_n} \frac{|dz|}{2\pi} \left| \frac{z_0 f(z)}{(z-z_0) z} \right| \le \frac{Az_0}{R_n - |z_0|} \to 0, \text{when } n \to \infty.$$ However, we also know that $$ \tag 7 I_n(z_0)-I_n(0) = f(z_0) - f(0) - \sum_{k=1}^n \left( \frac{1}{z_0-z_k} + \frac{1}{z_k} \right), $$ which taking the limit $n\to \infty$ gives the wanted result (1).

For an example of application of (1) see e.g. here.

$\endgroup$

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.