Take $f$ to be a holomorphic function which is analytic in the unit disk except at zero. Suppose we also have functions $\varphi_j$ that are holomorphic in the entire unit disk and such that there exists $n \geq 0$ $$[f(z)]^n + \varphi_{n-1}(z) [f(z)]^{n-1} + \cdots + \varphi_1(z) f(z) + \varphi_0(z) \equiv 0,$$ for every $z$ in the unit disk except zero. Is it possible to show that $f$ cannot have a pole or essential singularity?

## 1 Answer

I think indeed $f$ cannot have a pole or an essential singularity. The case $n=0$ cannot happen, so below say $n \geq 1$.

If $f$ had a pole of order $k>0$, then $f(z)^n$ has a pole of order $kn$, whereas the remaining terms $\varphi_{n-1}(z)f(z)^{n-1}+ \ldots + \varphi_0(z)$ can have at most a pole of order $k(n-1)$, so they cannot cancel the pole of $f(z)^n$.

Now assume $f$ has an essential singularity. By the Theorem of Casorati-Weierstrass, this is equivalent to requiring that the image of arbitrarily small punctured discs around $0$ is dense in $\mathbb{C}$. In other words for all $r,\epsilon>0$, $u \in \mathbb{C}$ there is $z$ with $0<|z|<r$ such that $|f(z)-u|<\epsilon$. The idea is that around $z=0$, the functions $\varphi_j$ are essentially constant, so the left hand side of the equality in your question essentially looks like plugging the function $f(z)$ into a polynomial of positive degree. Then since such a polynomial is surjective (as a function $\mathbb{C} \to \mathbb{C}$) and since the image of $f$ around $0$ is dense in $\mathbb{C}$, the image of the composition is still dense in $\mathbb{C}$, thus certainly not equal to $\{0\}$. If you want to make this precise, you need to play around with some $\epsilon$ and $\delta$ to make sure that the coefficient functions $\varphi_j$ are sufficiently close to their value at $z=0$.