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Is there a meromorphic function on $\mathbb{C}\setminus\{0\}$ which have poles in $z_n=\frac{1}{n}$ of the order $n$?

I have tried something with the theorem of Mittag-Leffler but that didn't help.

Have you an idea?

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  • $\begingroup$ Maybe I don't get the question: $\frac1{(z-1/n)^n}$? $\endgroup$ – Olivier Oloa Jun 7 '16 at 19:51
  • $\begingroup$ @OlivierOloa We want one meromorphic function that has the given poles. $\endgroup$ – zhw. Jun 7 '16 at 20:40
  • $\begingroup$ But the function of OlivierOloa hat the given poles haven't it? $\endgroup$ – user337060 Jun 7 '16 at 20:43
  • $\begingroup$ @N.Sch No, that function has only one pole. We want one function that has a pole of order n at each 1/n $\endgroup$ – zhw. Jun 7 '16 at 22:11
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Sure: Mittag-Leffler says that there exists a meromorphic function $m(z)\in \mathcal M(\mathbb C^*)$ on $\mathbb C^*=\mathbb C\setminus \{0\}$ with principal part (=polar part) at $\frac 1n$for example equal to, for example, $(z- \frac 1n)^{-n}$ .
Edit Of course we must use a version of Mittag-Leffler valid for an arbitrary domain in $\mathbb C$.
One can find it in Rudin's Real and complex analysis, Theorem 15.13.

But: The function $m(z)$ is guaranteed to have an essential singularity at $z=0$.
In other words it does not extend to a meromorphic function on $\mathbb C$.
Said yet differently, the restriction map $\mathcal M(\mathbb C)\to \mathcal M(\mathbb C^*)$ does not contain $m(z)$ in its image.

Generalizations:
Given an arbitrary open subset $U\subset \mathbb C$, an arbitrary discrete closed subset $D\subset U$ and an arbitrary principal part $m_d(z)$ at each $d\in D$ there exists a meromorphic function $m(z)\in \mathcal M(U)$ with principal part $m_d(z)$ at $d$.
Behnke-Stein proved that this result is still true if $U$ is an abstract, completely arbitrary non-compact Riemann surface.

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  • $\begingroup$ Thank you, so the answer is no? But the essential singularity at $z=0$ is not simportant since we are in $\mathbb{C}\setminus\{0\}$, is it? $\endgroup$ – user337060 Jun 7 '16 at 21:25
  • $\begingroup$ No, the answer is yes. $\endgroup$ – zhw. Jun 7 '16 at 21:27
  • $\begingroup$ The answer is yes there is such a meromorphic function on $\mathbb C^*$. You are right that if you stay in $\mathbb C^*$ the essential singularity at $0$ is irrelevant. So Mittag-Leffler does help, but you must use a version of Mittag-Leffler valid on an arbitrary domain of $\mathbb C$. But that's not a problem:such a version exists! $\endgroup$ – Georges Elencwajg Jun 7 '16 at 21:31
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Using the Weierstrass factorization theorem, one can construct an entire function $f(z)$ such that $f$ has a zero of order $n$ at each positive integer $n$ (and no other zeros). Then $$ g(z)=\frac{1}{f(\frac{1}{z})}$$ is meromorphic on $\mathbb{C}\setminus\{0\}$ with a pole of order $n$ at $\frac{1}{n}$ for all $n\geq 1$.

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I'd wager a pint of excellent red ale that the following $f$ does the job:

$$f(z) =\sum_{n=1}^{\infty} \frac{1}{n!(z-1/n)^n}.$$

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