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,
 A: 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.
