# How can I show that this meromorphic function is a rational function of two polynomials?

Here's my updated attempt:

Write$$f(z) = \sum_{n=-1}^{\infty} a_n(z-z_1)^n + ...+\sum_{n=-1}^{\infty} m_n(z-z_m)^n+\sum_{n=+1}^{-\infty} \psi_n(z)^n$$

with the last series being an expansion about the origin, and with $\large z=\frac{1}{w}$.

Now I subtract off the principal part of each Laurent series, getting

$$g:=\sum_{n=-1} a_n(z-z_1)^n + ...+\sum_{n=-1} m_n(z-z_m)^n - \frac{a_{-1}}{z-z_1} - ... - \frac{m_{-1}}{z-z_m}- {\psi_{1}}{z}$$

Now $g$ becomes entire and bounded, hence constant by Liouville's Theorem.

Moving the principal parts that I subtracted over to the L.H.S. gives me

$$M + \frac{a_{-1}}{z-z_1} + ... + \frac{m_{-1}}{z-z_m}- {\psi_{1}}{z}= \sum_{n=-1}^{\infty} a_n(z-z_1)^n + ...+\sum_{n=-1}^{\infty} m_n(z-z_m)^n+\sum_{n=+1}^{-\infty} \psi_n(z)^n$$

$$=f(z)$$

So, I have shown that this meromorphic function is a rational function, as required.

What do you think? Thanks,

The problem statement is:

Let $f(z)$ be analytic in the whole complex plane apart from simple poles at $z_1,…,z_m$ and let $f(\large \frac{1}{z})$ have a simple pole at $z=0$.

Part 1: Show that $f$ is a rational function $\large \frac{Q(z)}{P(z)}$ of polynomials $Q$ and $P$.

Part 2: What can you say about the degrees of $Q$ and $P$?

My work:

We know that $g(z)=(z-z_1)...(z-z_m)f(z)$ is an entire function.

This seems like a job for Liouville's Theorem, but I'm not sure how to utilize it right now, since I don't have a boundedness condition.

How could I use the assumption that $f(\frac{1}{z})$ has a simple pole at $z=0$?

I know that this means $f(z)$ has a simple pole at $\infty$, but this doesn't seem to help. I feel that this is the most important assumption in the question...

Any suggestions are welcome.

Thanks,

Subtract off the principal part of each pole, including the one at $\infty$, and you have a bounded entire function, which by Liouville is constant. What you subtracted was a rational function...
• Do you mean to subtract off the principal part of $m$ different Laurent series, each series being an expansion about one of the poles? Thanks @RobertIsrael, – User001 Dec 1 '15 at 8:55
• Yes, the principal part at $z=a$ of a function (analytic in a deleted neighbourhood of $a$) is the principal part of the Laurent series about $z=a$. After subtracting that, you have a function whose singularity at $z=a$ is removable; all the other singularities are the same as for the original function. – Robert Israel Dec 1 '15 at 16:39
• If $f(1/w) = c/w + \ldots$ as $w \to 0$, writing $z = 1/w$ gives you $f(z) = c z + \ldots$. The principal part at $\infty$ is then $c z$. – Robert Israel Dec 2 '15 at 3:42
• Your notation is not good, since the $c_n$ for different $m$ are different. – Robert Israel Dec 2 '15 at 3:43
As $f(\frac1z)$ has no zeroes near $z=0$, the set of zeroes of $f$ is bounded, hence finite, say (with multiplicity) they are $w_1,\ldots,w_n$. Then $$h(z)=\frac{(z-z_1)\cdots (z-z_m)}{(z-w_1)\cdots (z-w_n)} f(z)$$ is entire, has no zeroes and $h(\frac1z)$ has a pole of order $m+1-n$ (if $m+1>n$) or a zero of order $n-1-m$ (if $m+1<n$) or a removable discontinuity (iff $n=m+1$)at $z=0$. As $h$ has no zeroes, we can also consider $\frac1{h(\frac1z)}$ which is entire except possibly a pole at $z=0$ (if $m+1<n$). At any rate, at least one of $h(\frac1z)$, $\frac1{h(\frac1z)}$ is entire, hence bounded near $z=0$. Correspondingly, one of $h(z)$, $\frac1{h(z)}$ is entire and bounded (near $\infty$ and hence throughout), hence constant. We conclude $$f(z)=c\cdot \frac{(z-w_1)\cdots (z-w_n)} {(z-z_1)\cdots (z-z_m)}$$ with $c\ne 0$ and $n=m+1$.
• Hi @HagenvonEitzen, would it be easier / valid if instead I could show that my $g$ is a polynomial? My $g$ I think should have a pole at infinity, since on the R.H.S., $f$ has a pole at infinity. I am a little unfamiliar with your technique, so I wonder whether I could still proceed in my current direction ... thanks – User001 Dec 1 '15 at 7:26