I'm learning complex analysis, specifically series applications, and need help with the following problem:
Let $f(z)$ be an entire function which is not a polynomial. Show that for every $c \in \mathbb{C}$ there exists a complex sequence $\{z_n\}$ with $z_n \to \infty$ such that $f(z_n) \to c$.
Since $f(z)$ is holomorphic in the whole complex plane, then $f(z)$ can be expanded in a Taylor series around $z_0 = 0$, i.e.
$$f(z) = \sum_{n = 0}^{\infty}a_nz^n.$$
I suspect the solution to this exercise has to do with the Casorati-Weierstrass theorem but I don't know how to apply it in this context. If $f(z)$ is not a polynomial, can we say something of interest about the Taylor coefficients $a_n$ of $f(z)$ and why?
In a previous exercise a hint was given to consider the series expansion of $g(z) = f(1/z)$. My guess is that, in order to apply the Casorati-Weierstrass theorem, $g(z)$ must have a essential singularity at $z_0 = 0$. I'd like to understand why this is so and possibly some help with the application of the Casorati-Weierstrass theorem.
Some hints/steps :
If an entire function has a non-essential singularity at $\infty$, then it is a polynomial. In order to prove this, consider $g(z)=f(\dfrac{1}{z})$, and compare power series expansions.
From (1), conclude that $\infty$ is an essential singularity of $f$.
Casorati-Weierstrass tells you that the image of a deleted neighbourhood of a essential singularity is dense in $\mathbb{C}$. Every set of the form $\{z:|z|>n\}$ is a deleted neighbourhood of $\infty$.
Pick a point $c\in \mathbb{C}$. Pick neighbourhoods $B(c,\dfrac{1}{n})$, and observe that for each $n\in \mathbb{N}$, $f(\{z:|z|>n\}\cap B(c,\dfrac{1}{n}) \neq \emptyset$.
Pick $z_n$ in $\{z:|z|>n\}$.
Alternatively, argue contrapositively. Suppose $\not \exists ~z_n\to \infty$ such that $f(z_n)\to c$. Consider $g(z)=\dfrac{1}{f(\frac{1}{z})-c}$. Show that $g$ is either meromorphic or holomorphic at $0$, and conclude.