Let p be a positive integer. Consider theclosed curve $$r(t)=e^{it}, 0\leq t< 2\pi$$​.

Let f be afunction holomorphic in $$\{z:|z|<R\}$$​ whereR >1. If f has a zero only at $$z_0,0<|z_0|<R$$​ ,and it is of multiplicity q , then $$\frac{1}{2\pi i}\int_r \frac{f'(z)}{f(z)}z^pdz$$ equals?​

Correct option is A

​$$Let \ p \text{ be a positive integer. Consider the closed curve } r(t) = e^{it}, \ 0 \leq t \leq 2\pi. \\\text{Let } f \text{ be a function holomorphic in } \{ z \mid |z| < R \} \text{ where } R > 1. \\\text{If } f \text{ has a zero only at } z_0, 0 < |z_0| < R, \text{ and it has multiplicity } q, \\\text{then the integral } \frac{1}{2\pi i} \int_{r} \frac{f'(z)}{f(z)} z^p dz \text{ equals:} \\f(z) = (z - z_0)^q g(z) \\f'(z) = q(z - z_0)^{q-1} g(z) + (z - z_0)^q g'(z) \\\frac{1}{2\pi i} \int_{r} \frac{f'(z)}{f(z)} z^p \, dz = \frac{1}{2\pi i} \int_{r} \left[ \frac{q(z - z_0)^{q-1} g(z)}{(z - z_0)^q g(z)} + \frac{(z - z_0)^q g'(z)}{(z - z_0)^q g(z)} \right] z^p \, dz \\= \frac{1}{2\pi i} \int_{r} \left[ \frac{q}{(z - z_0)} z^p + \frac{g'(z)}{g(z)} z^p \right] dz \\\text{By the residue theorem, the integral simplifies to} \\= \text{Res}_{z=z_0} \left[ \frac{q}{z - z_0} z^p \right] = q z_0^p.$$​​