Cauchy's Integral Formula

Theorem
Let $D = \left\{{z \in \C : \left|{z}\right| \leq r}\right\}$ 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 $D$:


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

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

By induction we can show the general result


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

for $k=1,2,3,...$