Question

Differentiate Cauchy’s formula (3.9) or (3.10) to get

$\mathbf{f}\mathbf{\text{'}}\left(\mathbf{z}\right)\mathbf{=}\frac{\mathbf{1}}{\mathbf{2}\mathbf{\pi i}}{\oint }_{\mathbf{C}}\frac{\mathbf{f}\left(\mathbf{w}\right)\mathbf{dw}}{{\left(\mathbf{w}\mathbf{-}\mathbf{z}\right)}^{\mathbf{2}}}$or$\mathbf{f}\mathbf{\text{'}}\left(\mathbf{a}\right)\mathbf{=}\frac{\mathbf{1}}{\mathbf{2}\mathbf{\pi i}}{\oint }_{\mathbf{C}}\frac{\mathbf{f}\left(\mathbf{z}\right)\mathbf{dz}}{{\left(\mathbf{z}\mathbf{-}\mathbf{\alpha }\right)}^{\mathbf{2}}}$

By differentiating n times, obtain

${\mathbf{f}}^{\left(\mathbf{n}\right)}\left(\mathbf{z}\right)\mathbf{=}\frac{\mathbf{n}\mathbf{!}}{\mathbf{2}\mathbf{\pi i}}{\oint }_{\mathbf{C}}\frac{\mathbf{f}\left(\mathbf{w}\right)\mathbf{dw}}{{\left(\mathbf{w}\mathbf{-}\mathbf{z}\right)}^{\mathbf{n}\mathbf{+}\mathbf{1}}}$or${\mathbf{f}}^{\left(\mathbf{n}\right)}\left(\mathbf{\alpha }\right)\mathbf{=}\frac{\mathbf{n}\mathbf{!}}{\mathbf{2}\mathbf{\pi i}}{\oint }_{\mathbf{C}}\frac{\mathbf{f}\left(\mathbf{z}\right)\mathbf{dz}}{{\left(\mathbf{z}\mathbf{-}\mathbf{a}\right)}^{\mathbf{n}\mathbf{+}\mathbf{1}}}$

Short answer

The value is as follows:

$f^{\left(n\right)}\left(z\right)=\frac{n!}{2\pi i}\oint \frac{f\left(w\right)}{{\left(w-z\right)}^{n+1}}dw$

Step-by-step solution

Introduction

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.

Considering the formula

$f\left(\alpha \right)=\frac{1}{2\pi i}\oint \frac{f\left(z\right)}{z-1}dz$ …… (1)

Note that both sides are function of $\alpha$,so differentiating both sides with respect to $\alpha$,

We get

$\begin{array}{l}f\text{'}\left(\alpha \right)=\frac{d}{d\alpha }\left(\frac{1}{2\pi i}\oint \frac{f\left(z\right)}{z-\alpha }dz\right)\\ =\frac{1}{2\pi i}\oint \frac{d}{d\alpha }\left(\frac{f\left(z\right)}{z-\alpha }\right)dz\\ =\frac{1}{2\pi i}\oint \frac{f\left(z\right)}{{\left(z-\alpha \right)}^2}dz\end{array}$

Hence,

$f\text{'}\left(\alpha \right)=\frac{1}{2\pi i}\oint \frac{f\left(z\right)}{{\left(z-\alpha \right)}^2}dz$

Differentiating both sides of this with respect to $\alpha$, we get

$\begin{array}{l}f\text{'}\text{'}\left(\alpha \right)=\frac{1}{2\pi i}\oint \frac{d}{d\alpha }\left(\frac{f\left(z\right)}{{\left(z-\alpha \right)}^2}\right)dz\\ =\frac{2}{2\pi i}\oint \frac{f\left(z\right)}{{\left(z-\alpha \right)}^3}dz\end{array}$

Differentiating both sides of this with respect to$\alpha$,we get

$\begin{array}{l}f\text{'}\text{'}\text{'}\left(\alpha \right)=\frac{1}{2\pi i}\oint \frac{d}{d\alpha }\left(\frac{f\left(z\right)}{{\left(z-\alpha \right)}^3}\right)dz\\ =\frac{2\times 3}{2\pi i}\oint \frac{f\left(z\right)}{{\left(z-\alpha \right)}^4}dz\end{array}$

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

$f^{\left(n\right)}\left(\alpha \right)=\frac{2\times 3\times ...\times n}{2\pi i}\oint \frac{f\left(z\right)}{{\left(z-\alpha \right)}^{n+1}}dz$

Hence,

$f^{\left(n\right)}\left(\alpha \right)=\frac{n!}{2\pi i}\oint \frac{f\left(z\right)}{{\left(z-\alpha \right)}^{n+1}}dz$

Differentiating the formula

Consider the following formula

$f\left(z\right)=\frac{1}{2\pi i}\oint \frac{f\left(w\right)}{w-z}dw$ ………. (2)

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

We get

$\begin{array}{l}f\text{'}\left(z\right)=\frac{d}{dz}\left(\frac{1}{2\pi i}\oint \frac{f\left(w\right)}{w-z}dw\right)\\ =\frac{1}{2\pi i}\oint \frac{d}{dz}\left(\frac{f\left(w\right)}{w-z}\right)dw\\ =\frac{1}{2\pi i}\oint \frac{f\left(w\right)}{{\left(w-z\right)}^2}dz\end{array}$

Hence,

$f^{\text{'}}\left(z\right)=\frac{1}{2\pi i}\oint \left(\frac{f\left(w\right)}{{\left(w-z\right)}^2}\right)dz$

Differentiating both sides of this with respect to z,

we get

$\begin{array}{l}f\text{'}\text{'}\left(z\right)=\frac{1}{2\pi i}\oint \frac{d}{dz}\left(\frac{f\left(w\right)}{{\left(w-z\right)}^2}\right)dw\\ =\frac{2\times 3}{2\pi i}\oint \frac{f\left(w\right)}{{\left(w-z\right)}^3}dw\end{array}$

Differentiating both sides of this with respect to z,

we get

$\begin{array}{l}f\text{'}\text{'}\text{'}\left(z\right)=\frac{1}{2\pi i}\oint \frac{d}{dz}\left(\frac{f\left(w\right)}{{\left(w-z\right)}^3}\right)dw\\ =\frac{2\times 3}{2\pi i}\oint \frac{f\left(w\right)}{{\left(w-z\right)}^4}dw\end{array}$

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

we get

$f^{\left(n\right)}\left(z\right)=\frac{2\times 3\times ...\times n}{2\pi i}\oint \frac{f\left(w\right)}{{\left(w-z\right)}^{n+1}}dw$

Hence,

$f^{\left(n\right)}\left(z\right)=\frac{n!}{2\pi i}\oint \frac{f\left(w\right)}{{\left(w-z\right)}^{n+1}}dw$