# Cauchy Integral formulas

The Cauchy Integral formulas for a complex valued function ${\displaystyle f(z)}$ which is analytic inside and on a simple closed curve ${\displaystyle C}$ in some region ${\displaystyle {\mathcal {R}}}$ of the complex ${\displaystyle z}$ plane, for a complex number ${\displaystyle a}$ inside ${\displaystyle C}$

${\displaystyle f^{(n)}(a)={\frac {n!}{2\pi i}}\int _{C}{\frac {f(z)}{(z-a)^{n+1}}}dz.}$

## Proof by induction:

### Proof of the ${\displaystyle n=1}$ case

We consider the case of ${\displaystyle n=1}$. We begin by using the Cauchy integral formula to formally define the derivative of ${\displaystyle f(z)}$

${\displaystyle \lim _{h\rightarrow 0}{\frac {f(a+h)-f(a)}{h}}=\lim _{h\rightarrow 0}{\frac {1}{2\pi i}}\int _{C}f(z){\frac {1}{h}}\left[{\frac {1}{(z-a-h)}}-{\frac {1}{z-a}}\right]\;dz}$

${\displaystyle =\lim _{h\rightarrow 0}{\frac {1}{2\pi i}}\int _{C}f(z){\frac {1}{h}}\left[{\frac {z-a-(z-a-h)}{(z-a)(z-a-h)}}\right]\;dz=\lim _{h\rightarrow 0}{\frac {1}{2\pi i}}\int _{C}f(z)\left[{\frac {1}{(z-a-h)(z-a)}}\right]\;dz}$

Multiplying and dividing by ${\displaystyle (z-a)}$ and adding and subtracting <mmath> h [/itex] in the numerator allows us to write

${\displaystyle =\lim _{h\rightarrow 0}{\frac {1}{2\pi i}}\int _{C}f(z)\left[{\frac {z-a-h+h}{(z-a-h)(z-a)^{2}}}\right]\;dz={\frac {1}{2\pi i}}\int _{C}{\frac {f(z)}{(z-a)^{2}}}\;dz+\lim _{h\rightarrow 0}{\frac {1}{2\pi i}}\int _{C}f(z)\left[{\frac {h}{(z-a-h)(z-a)^{2}}}\right]\;dz.}$

We may estimate the second integral noting that ${\displaystyle |z-a-h|\geq |z-a|-|h|}$ and considering deforming the contour of integration to a circle of radius ${\displaystyle R}$ such that ${\displaystyle C:|z-a|=R}$ and from the maximum modulus theorem that ${\displaystyle |f(z)|\leq M}$

${\displaystyle \lim _{h\rightarrow 0}\left|{\frac {1}{2\pi i}}\oint _{C}{\frac {f(z)}{(z-a-h)(z-a)^{2}}}\;dz\right|<\lim _{h\rightarrow 0}{\frac {|h|}{2\pi }}{\frac {2\pi |z-a|M}{||z-a|-|h|||z-a|^{2}}}\leq \lim _{h\rightarrow 0}{\frac {M|h|}{\left|1-\displaystyle {\frac {h}{R}}\right|R^{2}}}=0,}$

proving the theorem for the case ${\displaystyle n=1}$.

It is possible to prove a finite number of cases using a similar construction.

### Proof going from ${\displaystyle k-1}$ to the ${\displaystyle k}$-th case

For the ${\displaystyle n=k}$-th case we can consider beginning with the ${\displaystyle n=k-1}$ case and formally constructing the derivative

${\displaystyle f^{(k)}(a)=\lim _{w\rightarrow a}{\frac {f^{(k-1)}(w)-f^{(k-1)}(a)}{w-a}}=\lim _{w\rightarrow a}{\frac {(k-1)!}{2\pi i}}\oint _{C}f(z){\frac {1}{w-a}}\left[{\frac {1}{(z-w)^{k}}}-{\frac {1}{(z-a)^{k}}}\right]\;dz}$

In the limit,

${\displaystyle \lim _{w\rightarrow a}{\frac {\displaystyle {\frac {1}{z-w}}-\displaystyle {\frac {1}{z-a}}}{w-a}}={\frac {\partial }{\partial a}}\left[{\frac {1}{(z-a)^{k}}}\right]={\frac {k}{(z-a)^{k+1}}}.}$

Thus,

${\displaystyle f^{(k)}(a)={\frac {k!}{2\pi i}}\oint _{C}{\frac {f(z)}{(z-a)^{k+1}}}\;dz.}$

We note that it is also prove the general case by differentiating each side of the Cauchy integral formula ${\displaystyle n}$ times with respect to ${\displaystyle a}$, where the ${\displaystyle n}$-th partial derivative with respect to ${\displaystyle a}$ is brought inside the integral

${\displaystyle {\frac {d^{n}f(a)}{da^{n}}}={\frac {1}{2\pi i}}\oint _{C}f(z){\frac {\partial ^{n}}{\partial a^{n}}}\left[{\frac {1}{(z-a)}}\right]\;dz={\frac {n!}{2\pi i}}\oint _{C}{\frac {f(z)}{(z-a)^{n+1}}}\;dz.}$

This is the complex integral form of the Leibniz rule for differentiating under the integral sign.