derivative quotient of holomorphic function as a contour integral

Question

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.

Answer 1

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.