1
$\begingroup$

We know that complex-analytic functions $f(z)$ agree with their power series representations on their domain of analyticity.

If a meromorphic function has simple poles at $z_1, ..., z_m$ and $\infty$, but is otherwise analytic on the whole complex plane, then does it agree with a sum of Laurent series expansions, with each series expanded about a simple pole of $f(z)$, $z_1, ... z_m$ and $\infty$ (simple pole at infinity), i.e. does

$$f(z) = \sum_{n=-1} c_n(z-z_1)^n + ...+\sum_{n=-1} c_n(z-z_m)^n$$

make sense?

Motivation: my main goal is to then subtract off the principal part of each Laurent series, getting

$$g:=\sum_{n=-1} c_n(z-z_1)^n + ...+\sum_{n=-1} c_n(z-z_m)^n - \frac{c_{-1}}{z-z_1} - ... - \frac{c_{-1}}{z-z_m}$$

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

$$ M + \frac{c_{-1}}{z-z_1} + ... + \frac{c_{-1}}{z-z_m}=\sum_{n=-1} c_n(z-z_1)^n + ...+\sum_{n=-1} c_n(z-z_m)^n $$

$$=f(z)$$

if my idea of decomposing the meromorphic function was correct. Then, I will have shown that this meromorphic function is a rational function.

Is this valid?

Thanks,

$\endgroup$
1
$\begingroup$

The representation $$ \tag 1 f(z) = \sum_{n=-1} c_n(z-z_1)^n + ...+\sum_{n=-1} c_n(z-z_m)^n $$ does not make sense. Apart from the formal problem that the coefficients $c_n$ are different for each Laurent series, every single Laurent series $$ \sum_{n=-1}^\infty c_n(z-z_j)^n $$ converges (only) in some disk (the largest disk with center $z_j$ which does not contain any other pole of $f$), and is not defined outside of that disk.

So each term in $(1)$ is defined on different domains and you can not add them.

But what you can do is to define $g$ as $$ g(z) = f(z) - \frac{c_{-1}^{(1)}}{z-z_1} - ... - \frac{c_{-1}^{(m)}}{z-z_m} $$ where $c_{-1}^{(j)}$ is the residue of $f$ at $z_j$, i.e. the coefficient of $(z-z_j)^{-1}$ in the Laurent series of $f$ at $z_j$.

From your assumption that $f$ has only simple poles it follows that $g$ is an entire function. But $g$ is not bounded because $f$ has a simple pole at $\infty$: $$ f(z) = az + O(1) \text { for } z \to \infty $$ for some $a \ne 0$. ($a$ is the residue of $f(1/z)$ at $z=0$.)

It follows that $$ g(z) = f(z) - \frac{c_{-1}^{(1)}}{z-z_1} - ... - \frac{c_{-1}^{(m)}}{z-z_m} - az $$ is entire and bounded, and therefore constant, and you have the representation $$ f(z) = \frac{c_{-1}^{(1)}}{z-z_1} + ... + \frac{c_{-1}^{(m)}}{z-z_m} + az + b $$ as a rational function.

The same can be done for arbitrary meromorphic function in the extended plane $\hat{\Bbb C}$ if you replace $$ \frac{c_{-1}^{(j)}}{z-z_j} $$ by the principal part of the Laurent series of $f$ at $z_j$, i.e. the part of the Laurent series with the (finitely many) negative exponents.

$\endgroup$
  • $\begingroup$ Hi @MartinR, thanks so much for your write-up - I was very happy to read it early this morning, and it had made perfect sense to me. But now that I am revisiting your solution it is confusing me just a little bit. There are two specific follow-up questions that I have, if you don't mind answering: 1) how can we just subtract off the principal parts of $f(z)$? I'm not sure why this is bothering me now, but here's my thoughts: in doing so, it would be a +infinity - infinity situation at each pole. $\endgroup$ – User001 Dec 3 '15 at 1:11
  • $\begingroup$ From elementary calculus, we learn that subtracting infinities is nonsense (while adding and multiplying +infinities results in +infinity again). So...isn't that sort of what we are doing here? Or, I might be missing the point with the arithmetic, and that it really is nothing more than subtracting off one term of $f$'s Laurent series at each pole. In the convergent annulus, the Laurent series agrees with $f$, so this should all make sense. Is this the correct way to think about it? And my second follow-up question is: $\endgroup$ – User001 Dec 3 '15 at 1:11
  • $\begingroup$ I notice that you expand around the origin with $az + O(1)$ ... why couldn't the Laurent series be $a_{-1}{1/w} + a_0 + a_1{w} + a_2{w^2}$ ... with z = 1/w? This expansion also has a simple pole at w = 0, i.e., at the point at infinity, but the remainder of the Laurent series is not in $O(1)$. Thanks for your time, @MartinR, $\endgroup$ – User001 Dec 3 '15 at 1:11
  • 1
    $\begingroup$ @LebronJames: Re #1: At each pole $z_j$, $f(z) - \frac{c_{-1}^{(j)}}{z-z_j}$ has a removable singularity as can be seen by expanding $f$ into a Laurent series in the neighbourhood of $z_j$. (That's what you probably meant in your second comment.) Therefore $g$ can be continued analytically to $z_j$. This works at each pole, therefore $g$ is entire. $\endgroup$ – Martin R Dec 3 '15 at 8:04
  • 1
    $\begingroup$ @LebronJames: Re #2: $f(z) = az + O(1)$ is the expansion of $f$ at infinity, not at the origin. You said that $f$ has a simple pole at infinity. $\endgroup$ – Martin R Dec 3 '15 at 8:06

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.