Let $f:\mathbb{C}\rightarrow \mathbb{C}$ be an entire function and let $ w\in \mathbb{C}$ be any point. We define the function $g(z)$ such that $g(z)=\frac{f(z)}{z-w}$
Find all the possible Laurent series expansions of the function g(z) centered around $z_0=w$
I noticed that the point $z_0=w$ is a pole for the function $g(z)$ however I am confused as to how to find the Laurent expansions of the function $g(z)$ dependent on f(z).
On
What would be the Laurent expansion of $f$ around $w$? Since you don't know the values of the coefficients for this expansion, you can introduce variables, for instance, $a_i$, as in
Let the entire function have the power series expansion, centered at $w$, $$ f(z) = \sum_{j=0}^{\infty} a_j (z-w)^j \text{} $$
Can you then write the Laurent expansion of $\frac{f(z)}{z-w}$ in terms of these coefficients?
Related: You seem to realize $\frac{f(z)}{z-w}$ has only one singularity, at $z = w$, so there will be only one Laurent expansion on the exterior of that point (i.e., on $\Bbb{C} \smallsetminus \{w\}$). You should be sure to say so.
You have to write the Luarent series expansion in terms of the power series for $f$. We can exapand $f(z)$ as $\sum\limits_{n=0}^{\infty} a_n(z-w)^{n}$ where $a_n=\frac {f^{(n)} (w)} {n!}$. So $g(z)=\frac {a_0} {z-w}+\sum\limits_{n=1}^{\infty} a_n(z-w)^{n-1}$. [Note that if $f(w)=0$ then $a_0=0$ and $g$ is also entire].
[To expand $g(z)$ in a power series around any point $z_0 \neq w$ you have to multiply two power series: One for $f(z)$ and one for $\frac 1 {z-w}$. The second series is $-\frac 1 {w-z_0} \sum\limits_{n=0}^{\infty} (\frac {z-z_0} {w-z_0})^{n}$].