Is my idea of decomposing a meromorphic function into a sum of Laurent series correct?

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,

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.

• 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. – User001 Dec 3 '15 at 1:11
• 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: – User001 Dec 3 '15 at 1:11
• 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, – User001 Dec 3 '15 at 1:11
• @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. – Martin R Dec 3 '15 at 8:04
• @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. – Martin R Dec 3 '15 at 8:06