Pole at Infinity implies Polynomial Function

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Let $f : \C \to \C$ be an entire function.

Let $f$ have a pole of order $N$ at $\infty$.

Then $f$ is a polynomial of degree $N$.


By Complex Function is Entire iff it has Everywhere Convergent Power Series, there exists a power series:

$\ds \map f z = \sum_{n \mathop = 0}^\infty a_n z^n$

convergent for all $z \in \C$, where $\sequence {a_n}$ is a sequence of complex coefficients.

This gives:

$\ds \map f {\frac 1 z} = \sum_{n \mathop = 0}^\infty \frac {a_n} {z^n}$

It is given that $\map f z$ has a pole of order $N$ at $\infty$, so $\map f {\dfrac 1 z}$ has a pole of order $N$ at $0$.

So $N$ is the least positive integer such that:

$\ds z^N \map f {\frac 1 z} = \sum_{n \mathop = 0}^\infty a_n z^{N - n}$

is holomorphic at $0$, with $a_N \ne 0$.

Therefore, all exponents of $z$, with non-zero coefficients, in this series must be non-negative.

So $a_n = 0$ for $n > N$.


$\ds \map f z = \sum_{n \mathop = 0}^N a_n z^n$

with $a_N \ne 0$.

That is, $f$ is a polynomial of degree $N$.