I have the function $f(z)$ that defined on the closed disc $ \bar D_3(0)$ by the integeral on the boundry $C_3(0)$:
$f(z) = \int_{C_3(0)} \frac{3w^2+7w+1}{w-z}dw $
I need to find $f'(1+i)$
Caushy integral theorem and it`s conclusions tells me that on $C_3(0)$ boundry, my function is:
$f(w) = 2 \pi i(3w^2+7w+1)$ and that the derivative of $f(z)$ is:
$f'(z)=\int_{ C_3(0)}\frac {(3w^2+7w+1)}{(w-z)^2}dw$ we use $\gamma:[0,1] \to \Bbb C $ defined $ \gamma (t)=3e^{2\pi it}$ to be our path, so we get:
$f'(1+i)=\int_{0}^{1}\frac {6 \pi ie^{2\pi i t}(3e^{4\pi it}+7e^{2\pi it}+1)}{(3e^{2\pi it}-(1+i))^2}dt$ as integral over a path is defined.
Is there a way to pass over this complicated integral? or there is an algebric trick with whom I`ll solve this integral?, or maybe somehow to choose another parametrization to my path?
You stated that $f(w) = 2\pi i (3w^2 + 7w + 1)$. Isn't it immediate that $f'(w) = 2 \pi i (6w + 7)$?