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I don't feel like I understand the concept of pole at infinity very well.

Suppose $f(z)$ has a pole of order $k$ at infinity. This means $f(z)/z^k$ is holomorphic in a neighborhood of infinity. Then, since $z$ and $1/z$ are both holomorphic in that neighborhood, we can use the fact that the product of holomorphic functions are holomorphic to conclude that $f(z)/z^n$ is holomorphic in that neighborhood for any $n$. So $f(z)$ has a pole of any order at infinity.

What's wrong with this line of logic?

Mika H.
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    $z$ is not holomorphic in a neighbourhood of infinity, it has a pole at $\infty$. Also, $f$ having a pole of order $k$ at infinity means that $f(z)/z^k$ is holomorphic in a neighbourhood of $\infty$ and the value at $\infty$ is not $0$. Perhaps best to concentrate on the function $h(z) = f(1/z)$. $f$ has a removable singularity/pole of order $k$/essential singularity at $\infty$ if and only if $h$ has a removable singularity/pole of order $k$/essential singularity in $0$. – Daniel Fischer Oct 19 '13 at 16:50

1 Answers1

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As Daniel Fischer said, the mistake comes from the assumption that $z$ is holomorphic at $\infty$. In fact, any function $f$ that is holomorphic on the whole Riemann sphere $\overline{\mathbb C}$ is constant. This is because the sphere is compact, hence $|f|$ must attain its maximum, but this implies $f$ is constant by the maximum principle.

If a function $f$ is holomorphic on $\mathbb C$ and not constant, then at $\infty$ it has either a pole or an essential singularity. More precisely, it has a pole at $\infty$ precisely when $f$ is a polynomial. For every entire function that is not a polynomial, $\infty$ is an essential singularity.

One more remark: a function is meromorphic on the Riemann sphere $\overline{\mathbb C}$ (i.e. has only poles as singularities) if and only if it is a rational function.

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