# $\varphi$ the Laurent series near pole $z_0$, how prove that $f-\varphi$ is analytic near $z_0$?

I'm reading the proof that a meromorphic function on the Riemann sphere must be a rational function, and I think I need to understand Laurent series better.

The idea of the proof I'm reading is that if $f$ is meromorphic on $\tilde{\mathbb{C}}$, then it has finitely many poles and we can subtract off the principal part of the Laurent series $\varphi_{i}$ at the poles $z_i$, then $f - \varphi_1 - \dotsb - \varphi_n -\varphi_\infty$ is analytic and bounded on $\varphi$.

How can we prove that if $\varphi_1$ is the principal part of the Laurent series of $f$ near the pole $z_1$, then $f-\varphi_1$ is analytic in a neighborhood of $z_1$? I think this should follow closely from the definition of the Laurent series.

A reference is sufficient, but I'm looking at the relevant sections of Ahlfors and Brown/Churchill and I feel like I'm getting lost in the epsilons.

• Eric: I think you mean to subtract off the principal part from the Laurent series of $f$ in a punctured disk around the respective poles. And, yes, when you subtract the principal part, you have an analytic function on the (unpunctured) disk. You could also consider clearing denominators at the poles by multiplying (rather than by subtracting). – Ted Shifrin Dec 4 '13 at 4:08
• @TedShifrin You're right, that's what I meant. Edited – Eric Auld Dec 4 '13 at 14:41

If $f$ has a pole at $z_{0}$, then, for some $R > 0$, one has $$f(z) = \sum_{n=-K}^{\infty}a_{n}(z-z_{0})^{n}=\frac{a_{-K}}{(z-z_{0})^{K}}+\cdots\frac{a_{-1}}{(z-z_{0})}+\sum_{n=0}^{\infty}a_{n}(z-z_{0})^{n},\;\;\; 0 < |z-z_{0}| < R.$$ Therefore $$g(z)=f(z) -\frac{a_{-K}}{(z-z_{0})^{K}}-\cdots-\frac{a_{-1}}{(z-z_{0})}=\sum_{n=0}^{\infty}a_{n}(z-z_{0})^{n},\;\;\; 0 < |z-z_{0}| < R,$$ has a power series expansion near $z=z_{0}$ and, thus, a removable singularity at $z_{0}$. There are only a finite number of singular terms and, so, convergence is not an issue. The singular terms at $\infty$ are positive powers of $z$. Once you subtract all singular terms from $f$, including those at $\infty$ (a polynomial), then you end up with a function which has an entire, bounded extension and, as such, must be a constant. Absorbing any non-zero constant into the polynomial gives the partial fraction expansion $$f(z) = p_{0}+p_{1}z+\cdots+p_{M}z^{M}+\sum_{l=1}^{L}\sum_{k=0}^{K_{l}}\frac{a_{l,k}}{(z-z_{l})^{k}}.$$
• You should say a pole at $z_0$, not near. – Ted Shifrin Dec 4 '13 at 4:04