# Meromorphic function with poles in $\frac{1}{n}$ of order $n$

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?

• Maybe I don't get the question: $\frac1{(z-1/n)^n}$? – Olivier Oloa Jun 7 '16 at 19:51
• @OlivierOloa We want one meromorphic function that has the given poles. – zhw. Jun 7 '16 at 20:40
• But the function of OlivierOloa hat the given poles haven't it? – user337060 Jun 7 '16 at 20:43
• @N.Sch No, that function has only one pole. We want one function that has a pole of order n at each 1/n – zhw. Jun 7 '16 at 22:11

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.

• 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? – user337060 Jun 7 '16 at 21:25
• No, the answer is yes. – zhw. Jun 7 '16 at 21:27
• 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! – Georges Elencwajg Jun 7 '16 at 21:31

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 $n\geq 1$.

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

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