Cauchy's Integral Formula

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Theorem

Let $D = \left\{{z \in \C : \left|{z}\right| \le r}\right\}$ 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.


General Result

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

for $k = 1, 2, 3, \ldots$


Proof

Also see


Source of Name

This entry was named for Augustin Louis Cauchy.