Find a sequence which uniformly converges f(z), and is of the form $\displaystyle\sum_{i=1}^{\infty} \frac{c_i}{w_i-z}$

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Let $f\colon\mathbb{C}\rightarrow \mathbb{C}$, analytic in the closed disk $D=\{z:|z-z_0|\leqslant R\}$.

Is there a way of defining a sequence of the form $\displaystyle\sum_{i=1}^{\infty} \frac{c_i}{w_i-z}$ such that $w_i \in \mathbb{C}$ \ $D$, $c_i \in \mathbb{C}$ and its partial sums uniformly converge $f(z)$?

Thoughts: I have no idea... except maybe for the riemannian sum of $\displaystyle f(z)=\int_\gamma \frac{1}{w-z}dw$ where $\gamma\colon t\mapsto z_0+Re^{2\pi it}$, but then $w_i\in D$...

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You can make your idea work. Note that the assumption is that $f$ is analytic on the closed disc $\{ |z - z_0| \le R \}$. This actually means (by a compactness argument) that there is some $R' > R$ such that $f$ is analytic on $\{ |z-z_0| < R' \}$. Then write $$ f(z) = \frac{1}{2\pi i}\int_{|w-z_0| = \rho} \frac{f(w)}{w-z}\,dw $$ where $R < \rho < R'$ and approximate the integral (uniformly) by Riemann sums.