Suppose that $f$ is analytic in $\{z \in \mathbb{C}: 0 < |z| < 1\}$ with essential singularity at $0$.

Question

Suppose that $f$ is analytic in $D^{*} = \{z \in \mathbb{C}: 0 < |z| < 1\}$ with essential singularity at 0.

Show that for each $w \in \mathbb{C}$, exists a sequence $\{z_n\} \in D^{*}$ such that $\displaystyle\lim_{n\to\infty}f(z_n) = w$.

I know that we can cover almost the whole complex plane as we approach to the essential singularity and Casoratti Weierstrass says that the closure of the image of f under the annulus is $\mathbb{C}$. The problem is how we construct that sequence!

I was trying to prove by contradiction, that is: for any sequence in $D^{*}$ exists an $\epsilon > 0$ such that $|f(z_n) - w| \geq \epsilon$. I did not find anything usefull.

Answer 1

Unless I misunderstand something, your proposition is false.

Consider $f(z) = e^{1/z}$, it has an essential singularity at $z=0$, but there is no $z_0$ such that $f(z_0)=0$.

---

Perhaps you are asked to prove

Show that for each $w\in\mathbb{C}$, exists a sequence $\{z_n\}\in D^\ast$ such that $f(z_n)\to w$.

In this case, this is a simple corollary of Weierstrass-Casorati.

---

Perhaps, alternatively, you are asked to prove

Show that for each $w\in\mathbb{C}$, with a single exceptional $w$, there exists a sequence $\{z_n\}\in D^\ast$ such that $f(z_n)= w$.

In this case, this is equivalent to great Picard theorem.