I have a trouble in the following problem:
Let $D=\left\{z\in\mathbb{C}~:~|z|=2\right\}$, and let $f:\mathbb{C}\setminus{D}\to\mathbb{C}$ be a function defined by $$f(\theta)=\int_{|z|=2}\frac{z^{3}+3z^{2}+z+1}{-2(z-\theta)^3}\,dz.$$ Find $f(0)$ and $f'(3i)$.
Is it possible to solve the above integration?
I have never seen kinds of integration, so I have no idea how to get started.
Give some advice! Thank you!
You can evaluate $f(0)$ directly from the definition of the integral. However if you know that the integral of the derivative of a function over a closed path is always 0 you can easily get $f(0)$ as $\int_{|z|=2} (-3/2) \frac 1 zdz=-3 \pi i$ since the other terms integrate to $0$. For $f'(3i)$ you don't have to calculate anything! By Cauchy's Theorem $f(\theta)=0$ whenever $|\theta| >2$ which implies $f'(\theta)=0$ whenever $|\theta| >2$, so the answer is $0$.