0
$\begingroup$

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?

$\endgroup$
2
$\begingroup$

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 $z=0$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.