2
$\begingroup$

It is known that for a holomorphic function $f$ over $\Omega$, if $C$ is a circle inside $\Omega$, then:
$f^{(n)}(z) = \frac{n!}{2\pi i}\int_C \frac{f(s)}{(s-z)^{n+1}} \, ds$

A textbook makes the following claim :
$f:\mathbb{D}\rightarrow\mathbb{C}$
where $\mathbb{D}$ is the unit disc, and $2f'(0) = \frac{1}{2\pi i}\int_{C_r} \frac{f(s)-f(-s)}{(s-z)^2} \, ds$

Here $C_r$ is a circle about the origin whose radius is $r$.
The second formula seems to follow easily from the first by setting $n=1$ and summing two identical integrals where a simple change of variable is made to one of the summands. Here is where it gets weird.

The text says that the second formula only holds whenever $0 \lt r \lt 1$. I see no reason for imposing $r \lt 1$. Why would the text go out of its way to make sure $r \neq 1$? I don't see what is so special about this case.

Additional Info: The statement is made in reference to an exercise problem. The text is Stein & Shakarchi: Princeton Lectures in Analysis II Complex Analysis. You can find the statement on page 65, in Chapter 2, exercise 7 where it says "Hint". At this time, the question and a solution can be found at https://6ba38298-a-62cb3a1a-s-sites.googlegroups.com/site/davegaebler/home-1/solution-manuals/SteinandShakarchiComplexAnalysis.pdf on page 5 of the pdf under "Exercise 7". The author of the solution also goes out of his way to assure $r \neq 1$ and uses a limit argument to compensate.

  • 0
    The link in the post is broken.2016-02-17

1 Answers 1

3

Note that in the statement of the Cauchy Integral Formula (which share the same necessary conditions as the resulting differentiation formulae), the closure of the disk $D$ must lie completely within the domain of holomorphicity $U.$ Since we don't know that $f$ is holomorphic on an open set containing the entire unit disk, the limit argument has to be used.