# Cauchy Integral formulas

The Cauchy Integral formulas for a complex valued function $f(z)$ which is analytic inside and on a simple closed curve $C$ in some region ${\mathcal {R}}$ of the complex $z$ plane, for a complex number $a$ inside $C$ $f^{(n)}(a)={\frac {n!}{2\pi i}}\int _{C}{\frac {f(z)}{(z-a)^{n+1}}}dz.$ ## Proof by induction:

### Proof of the $n=1$ case

We consider the case of $n=1$ . We begin by using the Cauchy integral formula to formally define the derivative of $f(z)$ $\lim _{h\rightarrow 0}{\frac {f(a+h)-f(a)}{h}}=\lim _{h\rightarrow 0}{\frac {1}{2\pi i}}\int _{C}f(z){\frac {1}{h}}\left[{\frac {1}{(z-a-h)}}-{\frac {1}{z-a}}\right]\;dz$ $=\lim _{h\rightarrow 0}{\frac {1}{2\pi i}}\int _{C}f(z){\frac {1}{h}}\left[{\frac {z-a-(z-a-h)}{(z-a)(z-a-h)}}\right]\;dz=\lim _{h\rightarrow 0}{\frac {1}{2\pi i}}\int _{C}f(z)\left[{\frac {1}{(z-a-h)(z-a)}}\right]\;dz$ Multiplying and dividing by $(z-a)$ and adding and subtracting <mmath> h [/itex] in the numerator allows us to write

$=\lim _{h\rightarrow 0}{\frac {1}{2\pi i}}\int _{C}f(z)\left[{\frac {z-a-h+h}{(z-a-h)(z-a)^{2}}}\right]\;dz={\frac {1}{2\pi i}}\int _{C}{\frac {f(z)}{(z-a)^{2}}}\;dz+\lim _{h\rightarrow 0}{\frac {1}{2\pi i}}\int _{C}f(z)\left[{\frac {h}{(z-a-h)(z-a)^{2}}}\right]\;dz.$ We may estimate the second integral noting that $|z-a-h|\geq |z-a|-|h|$ and considering deforming the contour of integration to a circle of radius $R$ such that $C:|z-a|=R$ and from the maximum modulus theorem that $|f(z)|\leq M$ $\lim _{h\rightarrow 0}\left|{\frac {1}{2\pi i}}\oint _{C}{\frac {f(z)}{(z-a-h)(z-a)^{2}}}\;dz\right|<\lim _{h\rightarrow 0}{\frac {|h|}{2\pi }}{\frac {2\pi |z-a|M}{||z-a|-|h|||z-a|^{2}}}\leq \lim _{h\rightarrow 0}{\frac {M|h|}{\left|1-\displaystyle {\frac {h}{R}}\right|R^{2}}}=0,$ proving the theorem for the case $n=1$ .

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

### Proof going from $k-1$ to the $k$ -th case

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

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

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

$f^{(k)}(a)={\frac {k!}{2\pi i}}\oint _{C}{\frac {f(z)}{(z-a)^{k+1}}}\;dz.$ We note that it is also prove the general case by differentiating each side of the Cauchy integral formula $n$ times with respect to $a$ , where the $n$ -th partial derivative with respect to $a$ is brought inside the integral

${\frac {d^{n}f(a)}{da^{n}}}={\frac {1}{2\pi i}}\oint _{C}f(z){\frac {\partial ^{n}}{\partial a^{n}}}\left[{\frac {1}{(z-a)}}\right]\;dz={\frac {n!}{2\pi i}}\oint _{C}{\frac {f(z)}{(z-a)^{n+1}}}\;dz.$ This is the complex integral form of the Leibniz rule for differentiating under the integral sign.