# Removable singularity at infinity implies the Laurent series has no positive powers

Suppose that f(z) is analytic on the complex plane minus a single point $$z_0$$. Suppose further that f has a simple pole at $$z_0$$ and a removable singularity at infinity. Prove that $$f(z) = \frac{A}{z − z_0} + B$$

Since $$z_0$$ is a simple pole, we know the Laurent series around $$z_0$$ is of the form $$f(z) = \frac{A}{z − z_0} + B + \sum_{n=1}^{\infty} a_n (z - z_0 )^n$$ where $$a_n = \frac{1}{2 \pi i}\int_C \frac{f(z)}{(z - z_0)^{n+1}} dz$$. And $$a_n = - Res_{w=0} \frac{1}{w^2}\frac{f(\frac{1}{w})}{(\frac{1}{w} -z_0)^{n+1}}$$ by the change of variable $$z = \frac{1}{w}$$. Since z = ∞ is a removable singularity of f(z), we also know $$lim_{x \to \infty} zf(\frac{1}{z}) = 0$$. However, I'm not sure if I am missing something or how to finish the argument that $$a_n = 0$$ for all $$n > 0$$

• You almost have it. Render $w=1/(\color{blue}{z-z_0})$. – Oscar Lanzi Jan 3 at 18:52
• Estimate your contour integral for $a_n$ where $C$ is a circle of radius $r$, using the fact that $f$ is bounded near $\infty$, and take $r \to \infty$. – Robert Israel Jan 3 at 19:03