​$$\text{For a positive integer } p, \text{ consider the holomorphic function:} \\f(z) = \frac{\sin z}{z^p}, \quad \text{for } z \in \mathbb{C} \setminus \{0\}.\text{For which values of } p \text{ does there exist}\\\text{ a holomorphic function } g: \mathbb{C} \setminus \{0\} \to \mathbb{C} \\\text{such that } f(z) = g'(z), \text{ for } z \in \mathbb{C} \setminus \{0\}?$$​​

Correct option is B

if , g'(z) = f(z), Then

$$\int{f(z)}=g(z)\\\text{f(z) is given as :}\\ f(z)=\frac{sin z}{z^p}\\f(z)=\frac{z-\frac{z^3}{3!}+\frac{z^5}{5!}\cdots\frac{z^{2n+1}}{(2n+1)!}}{z^p}\ (\text{expanding sin z} )\\Now,\ Let,\ p=2\\\text{Then f(z) becomes,}\\f(z)=\frac{sin z}{z^p}\\f(z)=\frac{z-\frac{z^3}{3!}+\frac{z^5}{5!}\cdots\frac{z^{2n+1}}{(2n+1)!}}{z^2}\\.\\\text{log(z) will be present in integration of f(z), so p=2 is not possible, same with p=4 and 6}\\i.e \textbf{Options A,C and D are incorrect}\\\text{Also, if p is odd then }\ logz\ \text{won't be present in integration of f(z)}\\ \textbf{Hence, p can be any odd number}$$​​