# 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 $a=2-i$. > $\begin{align} > \frac{1}{2\pi i}\oint_{\gamma}\frac{f(z)}{z-(2-i)}\;dz&=f(2-i) \\ > \frac{1}{2\pi i}\oint_{\gamma}\frac{f(z)}{z-(2-i)}\;dz&=3e^{2-i} \\ > \therefore\oint_{\gamma}\frac{3e^{z}}{z-2+i}\;dz&=2\pi i\cdot3e^{2-i} > \end{align}$