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Let $f_n$ be continuous on the open set $U$ and let $f_n \rightarrow f$ uniformly on compact sets. and if $U \supseteq \{z_n\}$ and $z_n \rightarrow z_0 \in U$ and $f_j$ are holomorphic, then what i want to show is for $0 < k \in \mathbb{Z}$, \begin{align} \left(\frac{\partial}{\partial z}\right)^k f_n(z_n) \rightarrow \left(\frac{\partial}{\partial z}\right)^kf(z_0) \end{align}

My trial is similar treatment for $\lim_{n\rightarrow \infty} f_n(z_n) = f(z)$. But i am not sure about considering derivatives.

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1 Answer 1

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Use Cauchy's integral formula. For $n$ so large that $\lvert z_n - z_0\rvert < r/2$, we have

$$\biggl(\frac{\partial}{\partial z}\biggr)^k f_n (z_n) = \frac{k!}{2\pi i} \int_{\lvert w - z_0\rvert = r} \frac{f_n(w)}{(w - z_n)^{k+1}}\,dw\tag{1}$$

if $r > 0$ is chosen so small that $\{ z : \lvert z - z_0\rvert \leqslant r\} \subset U$. Then argue why you can take the limit inside the integral in $(1)$.

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