# Cauchy's Integral Formula **A function that is [[Holomorphic Function|holomorphic]] on a disk can be completely determined by its values on the boundary of the disk.** > [!NOTE] Cauchy's Integral Formula > Suppose $S$ is an open subset of $\mathbb{C}$ and $f:S\rightarrow\mathbb{C}$ is a [[holomorphic function]]. > > **Cauchy's integral formula** states that if $D$ is a closed disk *completely contained* in $S$ and $\gamma$ is a *counterclockwise orientated circle* forming the boundary of $D$, then for any point $a$ in the [[interior]] of $D$, > $f(a)=\frac{1}{2\pi i}\oint_{\gamma}\frac{f(z)}{z-a}\;dz$ Cauchy's integral formula expresses values of functions which are holomorphic on a closed disk in terms of a [[contour integral]] around the boundary of the disk. Thus, the values of the function within the disk can be determined knowing only the values of the function *on the boundary*. As holomorphic functions are also all complex [[Analytic Function|analytic]] functions, they are infinitely differentiable and their derivatives of any order can be expressed in a similar integral. > [!NOTE] Cauchy's integral formula for differentiation > Suppose $S$ is an open subset of $\mathbb{C}$ and $f:S\rightarrow\mathbb{C}$ is a holomorphic function. > > **Cauchy's integral formula for differentiation**, or **Cauchy's differentiation formula**, states that if $D$ is a closed disk *completely contained* in $S$ and $\gamma$ is a *counterclockwise orientated circle* forming the boundary of $D$, then for any point $a$ in the interior of $D$, > $f^{(n)}(a)=\frac{n!}{2\pi i}\oint_{\gamma}\frac{f(z)}{(z-a)^{n+1}}\;dz$ ## Examples > [!NOTE] Example > Let $\gamma$ be a circle centred at $1-2i$ of radius $4$, traversed counterclockwise. > > Suppose the integral > $\oint_{\gamma}\frac{3e^{z}}{z-2+i}\;dz$ > is to be determined. > > Let $f(z)=3e^{z}$, which is holomorphic on the entire complex plane. As the point $2-i$ is in the interior of the disk bounded by $\gamma$, Cauchy's integral formula can be used where $a=2-i$. > $\begin{align} > \frac{1}{2\pi i}\oint_{\gamma}\frac{f(z)}{z-(2-i)}\;dz&=f(2-i) \\ > \frac{1}{2\pi i}\oint_{\gamma}\frac{f(z)}{z-(2-i)}\;dz&=3e^{2-i} \\ > \therefore\oint_{\gamma}\frac{3e^{z}}{z-2+i}\;dz&=2\pi i\cdot3e^{2-i} > \end{align}$