91Ó°ÊÓ

The Cauchy’s formula defined as the values of a holomorphic function inside a disk, which are determine by the values of the function on the boundary of disk.

$f\left(\mathrm{Î\pm }\right)=\frac{1}{2\mathrm{Ï€}i}âˆ\circledR \frac{f\left(z\right)}{z-1}dz$ …… (1)

Note that both sides are function of $\mathrm{Î\pm }$,so differentiating both sides with respect to $\mathrm{Î\pm }$,

We get

$\begin{array}{l}f\text{'}\left(\mathrm{Î\pm }\right)=\frac{d}{d\mathrm{Î\pm }}\left(\frac{1}{2\mathrm{Ï€}i}âˆ\circledR \frac{f\left(z\right)}{z-\mathrm{Î\pm }}dz\right)\\ =\frac{1}{2\mathrm{Ï€}i}âˆ\circledR \frac{d}{d\mathrm{Î\pm }}\left(\frac{f\left(z\right)}{z-\mathrm{Î\pm }}\right)dz\\ =\frac{1}{2\mathrm{Ï€}i}âˆ\circledR \frac{f\left(z\right)}{{\left(z-\mathrm{Î\pm }\right)}^2}dz\end{array}$

Hence,

$f\text{'}\left(\mathrm{Î\pm }\right)=\frac{1}{2\mathrm{Ï€}i}âˆ\circledR \frac{f\left(z\right)}{{\left(z-\mathrm{Î\pm }\right)}^2}dz$

Differentiating both sides of this with respect to $\mathrm{Î\pm }$, we get

$\begin{array}{l}f\text{'}\text{'}\left(\mathrm{Î\pm }\right)=\frac{1}{2\mathrm{Ï€}i}âˆ\circledR \frac{d}{d\mathrm{Î\pm }}\left(\frac{f\left(z\right)}{{\left(z-\mathrm{Î\pm }\right)}^2}\right)dz\\ =\frac{2}{2\mathrm{Ï€}i}âˆ\circledR \frac{f\left(z\right)}{{\left(z-\mathrm{Î\pm }\right)}^3}dz\end{array}$

Differentiating both sides of this with respect to$\mathrm{Î\pm }$,we get

$\begin{array}{l}f\text{'}\text{'}\text{'}\left(\mathrm{Î\pm }\right)=\frac{1}{2\mathrm{Ï€}i}âˆ\circledR \frac{d}{d\mathrm{Î\pm }}\left(\frac{f\left(z\right)}{{\left(z-\mathrm{Î\pm }\right)}^3}\right)dz\\ =\frac{2×3}{2\mathrm{Ï€}i}âˆ\circledR \frac{f\left(z\right)}{{\left(z-\mathrm{Î\pm }\right)}^4}dz\end{array}$

Hence, we see that after differentiating equation (1) for n times, we get

$f^{\left(n\right)}\left(\mathrm{Î\pm }\right)=\frac{2×3×...×n}{2\mathrm{Ï€}i}âˆ\circledR \frac{f\left(z\right)}{{\left(z-\mathrm{Î\pm }\right)}^{n+1}}dz$

Hence,

$f^{\left(n\right)}\left(\mathrm{Î\pm }\right)=\frac{n!}{2\mathrm{Ï€}i}âˆ\circledR \frac{f\left(z\right)}{{\left(z-\mathrm{Î\pm }\right)}^{n+1}}dz$

Consider the following formula

$f\left(z\right)=\frac{1}{2\mathrm{Ï€}i}âˆ\circledR \frac{f\left(w\right)}{w-z}dw$ ………. (2)

Note that both sides are function ofso differentiating both sides with respect to z,

We get

$\begin{array}{l}f\text{'}\left(z\right)=\frac{d}{dz}\left(\frac{1}{2\mathrm{Ï€}i}âˆ\circledR \frac{f\left(w\right)}{w-z}dw\right)\\ =\frac{1}{2\mathrm{Ï€}i}âˆ\circledR \frac{d}{dz}\left(\frac{f\left(w\right)}{w-z}\right)dw\\ =\frac{1}{2\mathrm{Ï€}i}âˆ\circledR \frac{f\left(w\right)}{{\left(w-z\right)}^2}dz\end{array}$

Hence,

$f^{\text{'}}\left(z\right)=\frac{1}{2\mathrm{Ï€}i}âˆ\circledR \left(\frac{f\left(w\right)}{{\left(w-z\right)}^2}\right)dz$

Differentiating both sides of this with respect to z,

we get

$\begin{array}{l}f\text{'}\text{'}\left(z\right)=\frac{1}{2\mathrm{Ï€}i}âˆ\circledR \frac{d}{dz}\left(\frac{f\left(w\right)}{{\left(w-z\right)}^2}\right)dw\\ =\frac{2×3}{2\mathrm{Ï€}i}âˆ\circledR \frac{f\left(w\right)}{{\left(w-z\right)}^3}dw\end{array}$

Differentiating both sides of this with respect to z,

we get

$\begin{array}{l}f\text{'}\text{'}\text{'}\left(z\right)=\frac{1}{2\mathrm{Ï€}i}âˆ\circledR \frac{d}{dz}\left(\frac{f\left(w\right)}{{\left(w-z\right)}^3}\right)dw\\ =\frac{2×3}{2\mathrm{Ï€}i}âˆ\circledR \frac{f\left(w\right)}{{\left(w-z\right)}^4}dw\end{array}$

Hence, we see that after differentiating equation (2) for n times,

we get

$f^{\left(n\right)}\left(z\right)=\frac{2×3×...×n}{2\mathrm{Ï€}i}âˆ\circledR \frac{f\left(w\right)}{{\left(w-z\right)}^{n+1}}dw$

Hence,

$f^{\left(n\right)}\left(z\right)=\frac{n!}{2\mathrm{Ï€}i}âˆ\circledR \frac{f\left(w\right)}{{\left(w-z\right)}^{n+1}}dw$