Show that there exists $z_n$ which converge to $z_0$ such that $f(z_n) \rightarrow a$ for an essential singularity

Question

Fix $a \in \Bbb C$ and suppose $f$ has an essential singularity at $z_0$. Show that there exists $z_n$ which converge to $z_0$ such that $f(z_n) \rightarrow a$

My work: Suppose $f$ has an isolated singularity at $z_0$. Let $f$ is analytic on punctured disk $\Delta^{*}(z_0,\epsilon)$ for some $\epsilon >0$ . Then for all $\delta < \epsilon$, $f(\Delta^{*}(z_0,\delta))$ is dense in $\Bbb C$. And so, we have $f(\Delta ^{*} (z_0,1/n)) \cap B(a,1/n)$ is non-empty whenever $1/n <\delta$. Pick a point $w_n$ in this. Then $|w_n-a|<1/n$ and $w_n$ is of the form $f(z_n)$ with $z_n \in \Delta ^{*} (z_0,1/n)$. Hence $f(z_n) \to a$ and $z_n \to z_0$.

Did miss anything? I appreciate your kind help. Thank you so much!

Answer 1

$f(\Delta (z_0,1/n)) \cap B(a,1/n)$ is non-empty whenever $1/n <\delta$. Pick a point $w_n$ in this. Then $|w_n-a|<1/n$ and $w_n$ is of the form $f(z_n)$ with $z_n \in \Delta (z_0,1/n)$. Hence $f(z_n) \to a$ and $z_n \to z_0$.

Answer 2

I see you are using Casorati-Weierstrass' theorem to prove the claim. You can construct $z_n$ inductively as follows:

Let $\delta_1=\varepsilon/2$. By Casorati-Weierstrass' theorem $f\big(\mathbb D(z_0,\delta_1)\setminus\{z_0\}\big)$ is dense in $\mathbb C$. In particular then this set contains some complex number in $\mathbb D(a,1)$, so we may pick $$ z_1\in\mathbb D(z_0,\delta_1)\setminus\{z_0\}\ \text{ with }\ |f(z_1)-a|<1. $$ Take $\delta_2=\min(|z_1-z_0|,\delta_1/2)$. The same argument as above but with $\delta_2$ produces $$ z_2\in\mathbb D(z_0,\delta_2)\setminus\{z_0\}\ \text{ with }\ |f(z_2)-a|<1/2. $$ Notice that $|z_2-z_0|<\delta_2\leq|z_1-z_0|$, so $z_2\neq z_1$. Take $\delta_3=\min(|z_2-z_0|, \delta_2/2)$. Again we get $$ z_3\in\mathbb D(z_0,\delta_3)\setminus\{z_0\}\ \text{ with }\ |f(z_3)-a|<1/3. $$ and $z_3\not\in\{z_1,z_2\}$ because $|z_3-z_0|<\delta_3\leq|z_2-z_0|\leq|z_1-z_0|.$

Iterating this process produces an infinite sequence $\{z_n\}_{n\in\mathbb N}$ in $\mathbb D(z_0,\varepsilon)\setminus\{z_0\}$ satisfying $$ z_n\in\mathbb D(z_0,\delta_n)\setminus\{z_0\}\ \text{ and }\ |f(z_n)-a|<1/n. $$ Clearly $f(z_n)\to a$. Moreover, by construction, we have $$ \delta_n\leq\frac{\delta_{n-1}}2\leq\frac{\delta_{n-2}}{2^2}\leq\ldots\leq\frac{\delta_1}{2^{n-1}}=\frac{\varepsilon}{2^n}\to0\ \text{ as }\ n\to\infty $$ and so $z_n\to z_0$ as $n\to\infty$.