Cauchy Integral Formula

Given a Contour Integral of the form

$$\oint_\gamma {f(z)\,dz\over z-z_0},$$ (1)

define a path $\gamma_0$ as an infinitesimal Circle around the point $z_0$ (the dot in the above illustration). Define the path $\gamma_r$ as an arbitrary loop with a cut line (on which the forward and reverse contributions cancel each other out) so as to go around $z_0$.

The total path is then

$$\gamma = \gamma_0+\gamma_r,$$ (2)

so

$$\oint_\gamma {f(z)\,dz\over z-z_0} = \oint_{\gamma_0} {f(z)\,dz\over z-z_0} + \oint_{\gamma_r} {f(z)\,dz\over z-z_0}.$$ (3)

From the Cauchy Integral Theorem, the Contour Integral along any path not enclosing a Pole is 0. Therefore, the first term in the above equation is 0 since $\gamma_0$ does not enclose the Pole, and we are left with

$$\oint_\gamma {f(z)\,dz\over z-z_0} = \oint_{\gamma_r} {f(z)\,dz\over z-z_0}.$$ (4)

Now, let $z \equiv z_0+re^{i\theta}$, so $dz = ire^{i\theta}\, d\theta$. Then

$$\oint_\gamma {f(z)\,dz\over z-z_0} = \oint_{\gamma_r}{f(z_0+re^{i\theta})\over re^{i\theta}} ire^{i\theta}\,d\theta$$

$$= \oint_{\gamma_r} f(z_0+re^{i\theta})i\,d\theta.$$ (5)

But we are free to allow the radius $r$ to shrink to 0, so

$$\oint_\gamma {f(z)\,dz\over z-z_0} = \lim_{r\to 0} \oint_{\gamma_r} f(z_0+re^{i\theta})i\, d\theta = \oint_{\gamma_r} f(z_0)i\,d\theta$$

$$= if(z_0)\oint_{\gamma_r}\,d\theta = 2\pi if(z_0),$$ (6)

and

$$f(z_0) = {1\over 2\pi i} \oint_\gamma {f(z)\,dz\over z-z_0}.$$ (7)

If multiple loops are made around the Pole, then equation (7) becomes

$$n(\gamma,z_0)f(z_0) = {1\over 2\pi i} \oint_\gamma {f(z)\,dz\over z-z_0},$$ (8)

where $n(\gamma,z_0)$ is the Winding Number.

A similar formula holds for the derivatives of $f(z)$,

$$f'(z_0) = \lim_{h\to 0} {f(z_0+h)-f(h)\over h}$$

$$= \lim_{h\to 0} {1\over 2\pi ih} \left({\,\oint_\gamma {f(z)\,dz\over z-z_0-h} -\oint_\gamma{f(z)\,dz\over z-z_0}}\right)$$

$$= \lim_{h\to 0} {1\over 2\pi ih} \oint_\gamma {f(z)[(z-z_0)-(z-z_0-h)]\,dz\over (z-z_0-h)(z-z_0)}$$

$$= \lim_{h\to 0} {1\over 2\pi ih} \oint_\gamma{hf(z)\,dz\over (z-z_0-h)(z-z_0)}$$

$$= {1\over 2\pi i} \oint_\gamma {f(z)\,dz\over (z-z_0)^2}.$$ (9)

Iterating again,

$$f''(z_0) = {2\over 2\pi i}\oint_\gamma{f(z)\,dz\over (z-z_0)^3}.$$ (10)

Continuing the process and adding the Winding Number $n$,

$$n(\gamma,z_0)f^{(r)}(z_0) = {r!\over 2\pi i}\oint_\gamma {f(z)\,dz\over(z-z_0)^{r+1}}.$$ (11)

See also Morera's Theorem

References

Arfken, G. ``Cauchy's Integral Formula.'' §6.4 in Mathematical Methods for Physicists, 3rd ed. Orlando, FL: Academic Press, pp. 371-376, 1985.

Morse, P. M. and Feshbach, H. Methods of Theoretical Physics, Part I. New York: McGraw-Hill, pp. 367-372, 1953.