For which $p$, does there exist an anti-derivative?

Question

For positive integer $p$, consider the holomorphic function

$f(z)=\frac{\sin (z)}{z^p}$ , for $z\in \mathbb{C}\setminus \{0\}$ .

For which values of $p$ , does there exist a holomorphic function $g:\mathbb{C}\setminus \{0\}\to \mathbb{C}$ such that $g'(z)=f(z)$ for $z\in \mathbb{C}\setminus \{0\}?$

$(1)$ All even integers .

$(2)$ All odd integers.

$(3)$ All multiples of $3$.

$(4)$ All multiples of $4$.

My thinking :- I am aware of the fact "Every analytic function on a simply connected domain has a primitive (anti-derivative)"

Here , the domain is not simply connected and I am not sure how else to proceed.

Please give a hint. Thank you.

Answer 1

Hint: Consider the series expansion of $\sin$. Note that monomial functions of the form $z \mapsto z^n$ have an anti-derivative (on $\Bbb C \setminus \{0\}$) iff $n \neq -1$.
(Convergence issues aside) This tells you that you need to pick a $p$ such that there is no $1/z$ term. After this, show that all such $p$ actually do give a function with an anti-derivative.