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In my book (in German) there is a theorem stated the following:

Every $f\in H(U)$ can be expressed as a power series, with $H(U)$ the set of all holomorphic functions on the subset $U\subset \mathbb{C}$. With $B_r(z_0)\subset U$ and $z\in B = B_p(z_0)(0<p<r)$ it follows: $$ f(z) = \sum_{n=0}^{\infty}a_n(z-z_0)^n,\quad a_n =\frac{1}{2\pi i} \int_{\partial B}^{} \frac{f(\zeta)}{(\zeta-z_0)^{n+1}}\, d\zeta $$

and also $$f^{(n)}(z) = \frac{n!}{2\pi i}\int_{\partial B}^{} \frac{f(\zeta)}{(\zeta-z)^{n+1}}\, d\zeta$$

I really don't understand how he gets this formula for the derivatives. When I define the $n$-th derivative on the "normal" way I get the following expression:

$$f^{(n)}(z) = \sum_{k=n}^{\infty}(k\cdot \ldots \cdot (k-n+1))\;a_k\;(z-z_0)^{k-n}$$

I just don't understand why this two expressions should be the same or how the author even got his formula(the third equation).

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    $\begingroup$ What book are you reading? $\endgroup$
    – user9464
    Feb 6, 2021 at 21:45
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    $\begingroup$ Folkmar Bornemann "Funktionentheorie". It's a german book. $\endgroup$
    – Sen90
    Feb 6, 2021 at 21:46
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    $\begingroup$ Presumably he takes the Cauchy Integral Formula $f(z)=\frac1{2\pi i}\int_{\partial B} \frac{f(w)}{z-w}dw$ and differentiates with respect to $z$ under the integral sign. $\endgroup$ Feb 6, 2021 at 21:48
  • $\begingroup$ So you are reading "Korollar 2.5.2"? $\endgroup$
    – user9464
    Feb 6, 2021 at 21:52
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    $\begingroup$ Yes. Under the corollary he birefly explains how he got (2.5.3b) with centering somehow(i haven't understood it). $\endgroup$
    – Sen90
    Feb 6, 2021 at 21:54

2 Answers 2

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First, for fixed $z\in B=B_\rho(z_0)$, the geometric series converges uniformly on $\partial B$: $$ \frac{1}{\zeta-z}=\sum_{n=0}^{\infty} \frac{\left(z-z_{0}\right)^{n}}{\left(\zeta-z_{0}\right)^{n+1}} $$

By the uniform convergence, you can switch the integral and summation sign if you apply the Cauchy integral formula (Theorem 2.5.1 in the book): $$ f(z)=\frac{1}{2 \pi i} \int_{\partial B} \frac{f(\zeta)}{\zeta-z} d \zeta =\frac{1}{2 \pi i} \int_{\partial B} {f(\zeta)} \sum_{n=0}^{\infty} \frac{\left(z-z_{0}\right)^{n}}{\left(\zeta-z_{0}\right)^{n+1}} d \zeta\\ =\sum_{n=0}^{\infty}\left(\frac{1}{2 \pi i} \int_{\partial B} {f(\zeta)} \frac{1}{\left(\zeta-z_{0}\right)^{n+1}} d \zeta\right)\left(z-z_{0}\right)^{n} \quad(z \in B) $$ the n-th coefficient of which gives you the formula: $$ f^{(n)}(z_0)=\frac{n !}{2 \pi i} \int_{\partial B} \frac{f(\zeta)}{(\zeta-z_0)^{n+1}} d \zeta\tag{1} $$

For $z\in B\setminus\{z_0\}$, the argument above shows that

$$ f^{(n)}(z)=\frac{n !}{2 \pi i} \int_{\partial B_\rho(z)} \frac{f(\zeta)}{(\zeta-z)^{n+1}} d \zeta\tag{2} $$

But Cauchy's theorem allows you to shift the contour to conclude that $$ f^{(n)}(z)=\frac{n !}{2 \pi i} \int_{\partial B_\rho(z)} \frac{f(\zeta)}{(\zeta-z)^{n+1}} d \zeta =\frac{n !}{2 \pi i} \int_{\partial B} \frac{f(\zeta)}{(\zeta-z)^{n+1}} d \zeta \tag{2'} $$


The original proof in your book.

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  • $\begingroup$ I know that last equation is true for $z_0$ but what is with $z \neq z_0$ ? I miss something... $\endgroup$
    – Sen90
    Feb 6, 2021 at 22:21
  • $\begingroup$ @Sen90 I don't understand your question. The long chains of identities for $f(z)$ is true for every $z\in B$. What do you mean by $z\ne z_0$? Which line are you referring to? $\endgroup$
    – user9464
    Feb 6, 2021 at 22:42
  • $\begingroup$ Sorry, but I am missing something... Could you please explain the last part of your answer beginning with "the n-th coefficient...." a little bit more. I don't understand where $f^{(n)}(z_0)=n!a_n$ comes into play. $\endgroup$
    – Sen90
    Feb 7, 2021 at 1:14
  • $\begingroup$ @Sen90: I think I see what you mean in the first comment. See my edits. $\endgroup$
    – user9464
    Feb 7, 2021 at 2:04
  • $\begingroup$ Exactly what I have missed. Thank you very much for your help. $\endgroup$
    – Sen90
    Feb 7, 2021 at 23:59
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Since$$f(z)=\frac1{2\pi i}\int_{\partial B}\frac{f(\zeta)}{\zeta-z}\,\mathrm d\zeta,$$you have\begin{align}f'(z)&=\frac1{2\pi i}\int_{\partial B}\frac{\mathrm d}{\mathrm dz}\left(\frac{f(\zeta)}{\zeta-z}\right)\,\mathrm d\zeta\\&=\frac1{2\pi i}\int_{\partial B}\frac{f(\zeta)}{(\zeta-z)^2}\,\mathrm d\zeta.\end{align}Then you have\begin{align}f''(z)&=\frac1{2\pi i}\int_{\partial B}\frac{\mathrm d}{\mathrm dz}\left(\frac{f(\zeta)}{(\zeta-z)^2}\right)\,\mathrm d\zeta\\&=\frac2{2\pi i}\int_{\partial B}\frac{f(\zeta)}{(\zeta-z)^2}\,\mathrm d\zeta,\end{align}and so on.

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