Computing nth order derivative of $\frac{1}{(1-z)}$ using Cauchy's Integral Formula

Question

Let $f(z)= \frac{1}{(1-z)}$ be a complex function. The derivatives of this function are $f^n (z) = \frac{n!}{(1-z)^{n+1}}$. I can see that this is true by computing the first order, second order, and third order derivatives and noticing the pattern, but can we prove $f^n (z) = \frac{n!}{(1-z)^{n+1}}$ using the extension of Cauchy's Integral Formula? This question arises from example 4 on page 194 of Brown/Churchill's Complex Variables and Applications.

Answer 1

It's follows from the Cauchy Integral formula:

$$f^{(n)}(z)=\frac{n!}{2\pi i}\int_C\frac{f(s)ds}{(s-z)^{n+1}}$$

Because what the formula means really, is that for any point $z=s_0$ enclosed by any closed countour $C$ we have:

$$f^{(n)}(z)|_{s_0}=\frac{n!}{2\pi i}\int_C\frac{f(s)ds}{(s_0-z)^{n+1}}$$

The Laurent series is exactly $\frac{1}{1-z}=f(z)$, so you have $s_0=1$ above and the residue is $R_{f}@(s_0=1)=1=numer(f)$, so the formula simplifies, precisely because the denominator now does not participate on the integration:

$$\begin{align} f^{(n)}(z)&=\frac{n!}{2\pi i}\int_C\frac{f(s)ds}{(1-z)^{n+1}}\\ &=\frac{n!}{2\pi i}\cdot\frac{1}{(1-z)^{n+1}}\cdot\int_C f(s)ds\\ &=\frac{n!}{2\pi i(1-z)^{n+1}}\cdot 2\pi i\cdot R_{f}@(s_0=1)\\ &=\frac{n!}{(1-z)^{n+1}} \end{align}$$