Cauchy's Integral Formula/General Result

Theorem
Let $D = \{z \in \C : |z| \leq r\}$ be the closed disk of radius $r$ in $\C$.

Let $f:U \to \C$ be holomorphic on some open set containing $D$.

Then for each $a$ in the interior of $D$ we have:


 * $\displaystyle f^{(n)}(a) = \frac {n!}{2\pi i}\int_{\partial D} \frac{f(z)}{(z-a)^{n+1} }\ dz$

Where $\partial D$ is the boundary of $D$, and is traversed anticlockwise.

Proof
By Cauchy's Integral Formula we have:


 * $\displaystyle f(a) = \frac 1{2\pi i}\int_{\partial D} \frac{f(z)}{(z-a)}\ dz$

so for $n = 0$ we are done. Now suppose that:


 * $\displaystyle f^{(n-1)}(a) = \frac {(n-1)!}{2\pi i}\int_{\partial D} \frac{f(z)}{(z-a)^{n} }\ dz$

We find that

Therefore we are done by induction.