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).
Best Answer
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.