On a remark regarding the Cauchy integral formula

[psie]

TL;DR

I'm reading about the Cauchy integral formula and I got stuck on a remark that comments on the fact that we can obtain a power series from this formula.

The way the formula is stated in my noname lecture notes is as follows:

Theorem 11 (Cauchy integral formula). If $f(z)$ is analytic in $\left\{\left|z-z_{0}\right|<R\right\}$ then for any $0<r<R$ we find $$f\left(z_{0}\right)=\frac{1}{2 \pi i} \int_{\left|z-z_{0}\right|=r} \frac{f(z)}{z-z_{0}} dz\tag1 $$ if we integrate in the counter clockwise direction.

Then they remark that:

This follows by translating the integral $(1)$ to $z_{0}=0$, writing $f(z) / z=a_{0} / z+g(z)$ with $a_{0}=f(0)$ and $g(z)$ is analytic in $\{|z|<R\}$ and using $\int_{|z|=1}\frac1{z}dz=2\pi i$. When $\left|z_{0}\right|<R$ then by expanding $\frac{1}{z-z_{0}}=\frac{1}{z} \cdot \frac{1}{1-z_{0} / z}=\frac{1}{z}\left(1+\frac{z_{0}}{z}+\left(\frac{z_{0}}{z}\right)^{2}+\ldots\right)$ in $(1)$ one obtains a power series expansion of $f\left(z_{0}\right)$ in $\left\{\left|z_{0}\right|<R\right\}$ by integrating.

The last sentence puzzles me deeply, specifically the part "When $|z_0|<R$...". Why can we expand $\frac1{z-z_0}$ when $|z_0|<R$? This makes little sense to me. I have noticed several typos in these notes, and maybe this is another one, but I'm not sure what the correct condition is.

 

[fresh_42]

I am a bit confused about that many discs. What we need is $|z_0|<|z|$ in order to use the geometric series for $|q|=\left|\dfrac{z_0}{z}\right|<1.$

Maybe my version of the technique is more convincing:
https://www.physicsforums.com/insig...ntiation-and-integration/#Complex-Integration

 

[mathwonk]

I'm just thinking about this confusing explanation in my head, but I believe he means something like: there is a power series in z0 that represents f(z0) at every z0 with |z0|<R. Namely, given such a z0, choose r with |z0| < r < R. Then every z in the integral formula will satisfy the condition fresh_42 points out is necessary, namely |z0| < |z|, ( since we are integrating over the circle |z| = r). Thus the series will converge to f(z0). Of course the coefficients in this series appear to depend on the choice of r, but they do not, since they also equal the derivatives of f at z=0.

I guess my argument needs the remark that the integral formula holds for any z0 with |z0| < r, not just at z0 = 0. I.e. you don't have to integrate around a circle actually centered at z0, any circle containing z0 and lying in the region of analyticity will do. This follows from the proof of the formula, in which you actually shrink the circle down around z0, noting that the integral does not change, so in the limits you get f(z0) times the "winding number " of the circle, times i, i.e. 2πi.f(z0).