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I was thinking about this situation: If $f(z)$ is holomorphic everywhere except at $a$ (where it has a pole), and has a removable singularity or a pole at $\infty$, does $(z-a)f(z)$ also have a removable singularity or a pole at $\infty$?

In the pole case, $f(z)/z^k$ is holomorphic in some neighborhood of $\infty$. Since $(z-a)$ is holomorphic, $(z-a)f(z)/z^k$ should also be holomorphic in that neighborhood of $\infty$. But can we conclude that? I'm not sure because $z-a$ is not defined at $\infty$.

In the removable singularity case, we can define $f(\infty)$ so that $f(z)$ is holomorphic in a neighborhood of $\infty$. Again, can we conclude anything about $(z-a)f(z)$?

Mika H.
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1 Answers1

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Yes, in that situation, $(z-a)f(z)$ has a pole at infinity, or, if $f$ had a zero at infinity, a removable singularity.

When $k$ is such that $f(z)/z^k$ is holomorphic in a neighbourhood of infinity, then

$$\frac{(z-a)f(z)}{z^{k+1}} = \frac{f(z)}{z^k} - a\frac{f(z)}{z^{k+1}}$$

also is holomorphic in a neighbourhood of $\infty$.

Daniel Fischer
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  • Thank you. This might be a basic fact, but how do you know that $f(z)/z^{k+1}$ is also holomorphic in a neighborhood of $\infty$? – Mika H. Oct 19 '13 at 15:56
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    If you use the characterisation/definition that $g$ is holomorphic in a neighbourhood of $\infty$ if and only if $h(z) = g(1/z)$ has a removable singularity at $0$, then it is readily seen that for $\frac{f(z)}{z^{k+1}}$, we get $h(w) = w^{k+1}f(1/w)$. If $w^kf(1/w)$ has a removable singularity at $0$ (i.e. if $\frac{f(z)}{z^k}$ is holomorphic at $\infty$), then $w^{k+1}f(1/w)$ also has a removable singularity at $0$ (in fact, a zero). – Daniel Fischer Oct 19 '13 at 16:01