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 was trying this:
By Casoratti-Weierstrass, for the image of $f$ under any neighborhood of $0$ is dense in $\mathbb{C}$. That is, any $\epsilon > 0$, $B(w,\epsilon)\cap f(D^{*})\neq \emptyset$
Then for each $w \in \mathbb{C}$, $\exists z_w \in D^{*}$ such that $f(z_w) \in B(w,\epsilon)$.
Can I say that $\exists z_w \in D^{*}$ such that $f(z_w) = w$?
I want that to construct the sequence as follows:
$z_n \in B_n := B(z_w,\frac{1}{n})\cap D^{*}$
Then $z_n \in D^{*}$ for all $n$ and $\lim_{n\to\infty}z_n = z_w$. Hence,
$$\lim_{n\to\infty}f(z_n) = f(z_w) = w$$
But I'm not sure I can do that... Any idea on how to complete this proof?
Fix $w\in \mathbb C.$ Then for each $n=1,2,\dots$ there exists $z_n \in D^*$ such that $|f(z_n)-w|<1/n.$ That's all you need. And no, there need not be any $z\in D^*$ with $f(z)=w.$ Take $w=0$ and $f(z)=e^{1/z}$ to see this.