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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.

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Could give the textbook and page number? –  Potato Jun 16 '12 at 1:25
    
I agree there is no good reason for imposing that restriction, unless there is some other condition the text has imposed. –  Potato Jun 16 '12 at 1:26
    
I added additional information. There was the additional condition that $f$ is holomorphic on the unit disc. –  Mark Jun 16 '12 at 1:58
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1 Answer 1

up vote 3 down vote accepted

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.

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