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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 $n > 0$

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    $\begingroup$ You almost have it. Render $w=1/(\color{blue}{z-z_0})$. $\endgroup$ – Oscar Lanzi Jan 3 at 18:52
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    $\begingroup$ 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 $r \to \infty$. $\endgroup$ – Robert Israel Jan 3 at 19:03

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