I am having a problem to prove this question.
Fix $a \in \mathbf{C}$ and suppose $f$ has an essential singularity at $z_0$. Show that there exist a sequence $z_n$ which converges to $z_0$ such that $f(z_n)\rightarrow a$
Since $f$ has an essential singularity at $z_0$, by the definition of an essential singularity $f$ is analytic in $\vartriangle^*(z_0, \epsilon)$ and $f (\vartriangle^*(z_0, \delta))$ is dense in $\mathbf{C}$, with $\delta< \epsilon$.
So what I am saying here is because of the density, we have $a\in Cl(f (\vartriangle^*(z_0, \delta)))$. But I am getting stuck since I am having no clue for $a$ except it is fixed.
So I would appreciate any help or hints for that. Thank you.
The definition of an essential singularity is that it is an isolated singularity that is not a pole or removable. That the image of a punctured disk is dense is a theorem:
You can use the Casorati–Weierstrass theorem repeatedly on smaller and smaller discs around $z_0$ to construct a sequence. To wit:
Let $(R_n)$ and $(r_n)$ be sequences of positive real numbers, both converging to $0$.
Let $D_1 = \{z: 0 <\lvert z-z_0 \rvert < r_1\}$. By the C–W theorem, there is $z_1 \in D_1$ with $\lvert f(z_1)-a \rvert < R_1 $.
Now let $D_2 = \{z: 0 <\lvert z-z_0 \rvert < r_2\} $. Again, by the C–W theorem, there is $z_2 \in D_2$ with $\lvert f(z_2)-a \rvert < R_2 $. (It may be the same as $z_1$. But eventually we will have an $r_k$ with $\lvert z_1 - z_0 \rvert > r_k$, so the sequence is not constant.)
We now repeat this process to construct the sequence $z_n$: we end up with a sequence $z_n$ with $0<\lvert z_n-z_0 \rvert < r_n \downarrow 0$, with $\lvert f(z_n)-a \rvert < R_n \downarrow 0$, so $z_n \to z_0 $ and $f(z_n) \to a$ as required. If desired, the sequence can be forced to have distinct terms simply by choosing the first $r_k$ so that none of the previous $z_n$ are in $\{z: 0 <\lvert z-z_0 \rvert < r_k\}$ at each stage.