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

Question

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,

Answer 1

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.