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I have forgotten much of my complex analysis, so I would appreciate some help with the following.

Suppose $f$ is holomorphic and $\Gamma$ is a circle of radius $r$ about $0$. Why is $$\frac{f(z)-f(0)}{z}=\frac{1}{2\pi i}\int_\Gamma \frac{f(s)}{(s-z)s}ds \text{, for } 0<|z|<r? $$

I suspect one shows it using Cauchy's formula, but I don't see how.

Thank you in advance.

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Hint: $\frac{1}{(s-z)s}=\frac{1}{z}\left(\frac{1}{s-z}-\frac{1}{s}\right)$. –  Micah Jan 30 '13 at 17:33
    
Great! I knew it was simple. –  Cantor Jan 30 '13 at 17:37
    
@Cantor: In addition to accepting an answer, you can vote on whatever answers you feel deserve a vote. –  robjohn Jan 30 '13 at 18:45

1 Answer 1

up vote 1 down vote accepted

Let $|z| < 1$, by Cauchy's integral formula:

$$ f(z) = \frac{1}{2\pi i} \int_\Gamma \frac{f(s)}{s-z}\,ds $$

Applied to $z = 0$ we get:

$$ f(0) = \frac{1}{2\pi i} \int_\Gamma \frac{f(s)}{s}\,ds $$

Thus:

$$ \frac{f(z) - f(0)}{z} = \frac{1}{2\pi i} \int_\Gamma \frac{f(s)}{z} \left(\frac{1}{s-z} - \frac{1}{s}\right) \,ds $$

Simplify to get the result you have.

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