- #1

psie

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- TL;DR Summary
- 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:

Then they remark that:

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.

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.