# Detailed proof that no essential singularity at infinity implies polynomial

Suppose $f(z)$ is holomorphic in the whole plane, and that $f(z)$ does not have an essential singularity at $\infty$. Prove that $f(z)$ is a polynomial.

I've tried following the hint given in this question. Since $f(z)$ has a nonessential singularity at $\infty$, so $g(z)=f(1/z)$ has a nonessential singularity at $0$. There are two cases:

1) $g(z)$ has a removable singularity at $0$. This means $\lim_{z\rightarrow 0}zg(z)=0$.

2) $g(z)$ has a pole at $0$. This means $g(z)=h(z)/z^k$ for some analytic function $h(z)$ such that $h(0)\neq 0$.

How can I finish each of these cases?

• In both cases, $z^k g(z)$ is analytic in some neighborhood of $0$. So $z^kg(z)$ has a power series expansion valid in some neighborhood of $0$. $f$ is holomorphic, so $f$ also has a power series around $0$. How are these two power series related? – Gyu Eun Lee Oct 19 '13 at 4:16
• We can write $f(z)=f(0)+f'(0)z+f''(0)z^2/2!+\ldots$. For $z^kg(z)$, the $i$th derivatives at $0$ vanishes for $i=1,2,\ldots,k-1$, and the $k$th derivative should be $k!g(0)$. How is this helpful? – PJ Miller Oct 19 '13 at 4:27
• $z^k g(1/z)=z^k f(z)$, and since the right side is holomorphic the left side is too. But we already know that $z^k g(z)$ is holomorphic at $0$, so $z^k g(z)$ has a power series expansion around $0$. So write $z^k g(z)$ has a power series, and then write $z^k g(1/z)$ in terms of this power series. This power series has terms of the form $z^{-n}$, yet it is holomorphic at $0$. How can this be? – Gyu Eun Lee Oct 19 '13 at 4:35

1) Since $g$ is continuous, we can bound $g(z)$ inside some interval of $z$. Since $g(z)$ is defined as $f(1/z)$, it turns out that from the bound on $z$ we deduce that $f$ is a constant function:

There exists $M, \varepsilon >0$ such that $|g(z)|\leq M$ for all $|z| \in (0,\varepsilon)$. Hence $|f(z)|\leq M$ for all $|z| > 1/\varepsilon$. Letting $\varepsilon \to \infty$ $f$ is bounded and entire. It follows by Liouville's Theorem that $f$ is a constant function.

2) We use that $g$ has a Laurent expansion at $0$ and $f$ has a Taylor expansion since $f$ is entire. We can 'invert' the Laurent expansion so to say and by uniqueness of the Laurent expansion see that $f$ must be a polynomial.

Since the pole is of order $m$, the Laurent expansion of $g$ at $0$ is $$g(z) = \sum_{k=-m}^{\infty} a_k z^k$$ for $|z|\in (0,\varepsilon)$. We can invert this to get $f$: $$f(z) = \sum_{k = -\infty}^{m} a_{-k}z^k$$ for $|z|>1/\varepsilon$. The Taylor expansion of $f$ around $0$ given by $$f(z)=\sum_{k=0}^{\infty}b_kz^k$$ and the Laurent expansion must be equal by uniqueness, so $$f(z)=\sum_{k=0}^{m} b_k z^k$$ where $a_{-k}=b_k$ for all $k$. So $f$ is a polynomial.

See the following pdf: math.berkeley.edu/~mjv/Math185hw8.pdf

• The above link is not active – Tutankhamun Aug 9 '18 at 14:54

Since we have a non-essential singularity at $$\infty$$ we have an $$h$$ such that $$z^{-h}f(z)$$ has a limit as $$z$$ tends to $$\infty$$ that is neither $$0$$ nor $$\infty$$. This means that $$f$$ cannot have more than $$h$$ zeroes at $$0$$ because otherwise $$f(z) = z^{h+k}f_k(z)$$ and hence limit of $$z^{-h}f(z)$$ would not be finite. This also means that $$f^{(h+1)}(0) \neq 0$$ (the $$h+1$$st derivative is not $$0$$). Now $$f(z) = f(0) + f'(0)z + ... + z^{h+1}g(z)$$ where $$g(z)$$ is analytic in all of complex plane (including $$0$$). Hence $$g(0)$$ is finite and $$\neq 0$$ (because $$g(0) = f^{h+1}(0)$$) Also observe that we can write: $$z^{-h}f(z) = z^{-h}(f(0) + f'(0)z + ... + z^{h+1}g(z))$$

Now since $$\lim_{z \to \infty} z^{-h}f(z)$$ exists we see that $$\lim_{z \to \infty} zg(z)$$ also exists.

Also observe that $$g(z)$$ cannot be $$0$$ at infinity because otherwise $$g(1/t) = t^{k}g_2(t) \implies g(z) = \frac{g_2(1/z)}{z^{k}}$$ and would mean $$g(0)$$ is not defined.

All this implies that $$g(z) = 0$$