Let $f\in \mathcal{O}(D=\mathbb{D}(0,1))$
What does it take to make sure that $g(w)=\int_{\partial D(0,r)} \frac{f'(z)}{f(z)-w}$ is well defined and holomorphic for $r<1$ and $w\in D(0,r')$ with $r'<r$?
What I did:
for $g$ to be well defined, I guess it is enough showing that $|f(z)-w| > 0$ for every $z\in D$ and $w\in D(0,r')$ and that $\frac{f'(z)}{f(z)-w}$ is continuous on $\partial D(0,r)$ .
For $g$ to be holomorphic, I am stuck, any help much appreciated.
This can be proved using Morera's theorem: https://en.wikipedia.org/wiki/Morera%27s_theorem
Let $\gamma: [0,1] \rightarrow D(0,r^{\prime})$ be a closed curve that does not intersect the boundary. Then, \begin{equation} \int_{\gamma}g(w)\,dw = \int_{\gamma}\int_{\partial D(0,r)}\frac{f^{\prime}(z)}{f(z)-w}\,dz \,dw = \int_{\partial D(0,r)}f^{\prime}(z)\int_{\gamma}\frac{1}{f(z)-w}\,dw \end{equation} Since $|w - f(z)| > 0$ for all $w \in D(0,r^{\prime})$ and $z \in $D(0,1)$\frac{1}{f(z)-w}$ is holomorphic on $\partial D(0,r)$ and therefore the inner integral is $0. Hence, by Morera's theorem the result follows.