Let  f  be a holomorphic function on the disc $$\{z \in \mathbb{C}: |z| < 2\}$$​ 

Assume that the only zero of  f  in the closed unit disc  $$\{z \in \mathbb{C}: |z| \leq 1\}$$​​

is a simple zero at the origin. Let  $$\gamma$$​ be the positively oriented circle  $$\{z \in \mathbb{C}: |z| = 1\}.$$​​

The integral  $$\int_\gamma \frac{dz}{f(z)}$$​  equals ?

Correct option is C

​$$\text{Let } f \text{ be a holomorphic function on the disc } \{z \in \mathbb{C}: |z| < 2\}. \\[10pt]\text{ Assume that the only zero of } f \text{ in the closed unit disc } \{z \in \mathbb{C}: |z| \leq 1\}\\[10pt] \text{ is a simple zero at the origin. Let } \gamma \text{ be the positively oriented circle } \{z \in \mathbb{C}: |z| = 1\}.\\[10pt] \text{Since } f \text{ has a simple zero at } z=0, \text{ we can write } f(z) = z g(z) \text{ where } g(z) \text{ is holomorphic and } g(0) \neq 0. \\[10pt]\text{Then } f'(z) = g(z) + z g'(z). \\[10pt]\text{At } z=0, \text{ we have } f'(0) = g(0) + 0 \cdot g'(0) = g(0). \\[10pt]\text{Now consider the integral } \int_\gamma \frac{dz}{f(z)}. \\[10pt]\text{We have } \frac{1}{f(z)} = \frac{1}{z g(z)}. \\[10pt]\text{We can use the Residue Theorem to evaluate the integral. The residue of } \frac{1}{f(z)} \text{ at } z=0 \text{ is given by } \\[10pt]\text{Res}\left(\frac{1}{f(z)}, 0\right) = \lim_{z \to 0} (z-0) \frac{1}{f(z)} = \lim_{z \to 0} z \frac{1}{z g(z)} = \lim_{z \to 0} \frac{1}{g(z)} = \frac{1}{g(0)}. \\[10pt]\text{Since } f'(0) = g(0), \text{ we have } \text{Res}\left(\frac{1}{f(z)}, 0\right) = \frac{1}{f'(0)}. \\[10pt]\text{By the Residue Theorem, } \int_\gamma \frac{dz}{f(z)} = 2\pi i \cdot \text{Res}\left(\frac{1}{f(z)}, 0\right) = 2\pi i \cdot \frac{1}{f'(0)} = \frac{2\pi i}{f'(0)}. \\[10pt]\text{Thus, the integral } \int_\gamma \frac{dz}{f(z)} \text{ equals } \frac{2\pi i}{f'(0)}. \\[10pt]\text{Final Answer: The final answer is C.}$$​​