I am stuck in a problem about holomorphic functions and using of Cauchy integral formula. I really have no idea how to start, so i would be glad if somebody could help me with it.
Let $C=C(0,1)$ a circle with center $0$ and radius $1$, $D=D_{1}(0)$. On $D$ through $z\rightarrow \frac{1}{2\pi i}\int_{C} \frac{dw}{w(w-z)}$ is defined a holomorphic function $f_{1}$ and on $\mathbb{C} \setminus \bar D$ is defined another holomorphic function $f_{2}$. Find $f_{1}$ and $f_{2}$. In which points $w\in \partial D$ is $\lim_{z\rightarrow w}f_{1}(z) = \frac{1}{w}$ or $\lim_{z\rightarrow w}f_{2}(z) = \frac{1}{w}$?
I know that $\frac{1}{w}$ is holomorphic. But how to get to these limes or to find the function. Thank you in advance!
Note that $$ \frac{1}{w (w-z)} = \frac{1}{z(w-z)} - \frac{1}{z w}.$$ If $|z| < 1$ and $z \neq 0$ then as a function of $w$ this has two poles on the unit disc. One at $w=0$ with residue $-1/z$ and one at $w = z$ with residue $1/z$. Since $f_1(z)$ is the sum of all residues in the unit disc this means that $f_1(z) = 1/z - 1/z = 0$, which also holds for $z=0$.
If on the other hand $|z| > 1$ then there is only one pole in the unit disc at $w = 0$ with residue $-1/z$. Therefore $f_2(z) = -1/z$.
Both $f_1$ and $f_2$ extend over $C$: $f_1$ is identically zero on $C$ and $f_2(w) = -1/w$ for $w \in C$. Neither are equal to $1/w$.