Proof of Cauchy Integral Formula

ZaidAlyafey

Well-known member
MHB Math Helper
I want to prove the Cauchy integral formula :

$$\oint_{\gamma} \, \frac{f(z)}{z-z_0}\,dz= 2\pi i f(z_0)$$

$$\text{so we will integrate along a circle that contains the pole .}$$

$$|z-z_0|= \delta \,\text{ which is a circle centered at the pole and has a radius }\delta\,\,$$

$$z =z_0+\delta e^{i\theta }\,\, so \,\,dz =i\delta e^{i\theta }\,d\theta$$

$$\text{We will make the radius as small as possible to contain the pole . }$$

$$\lim_{\delta \, \to 0 } \, \oint_{0}^{2\pi} \, \frac{f(z_0 + \delta \, e^{i\theta })\cdot i\delta e^{i\theta } }{z_0+ \delta \, e^{i\theta }-z_0}\,d\theta$$

$$\lim_{\delta \, \to 0 }\, \oint_{0}^{2\pi} \, \frac{f(z_0 + \delta \, e^{i\theta })\cdot i\delta e^{i\theta } }{\delta \, e^{i\theta }}\,d\theta$$

$$\lim_{\delta \, \to 0 }\,i \oint_{0}^{2\pi} \,f(z_0 + \delta \, e^{i\theta }) \,d\theta$$

$$i \oint_{0}^{2\pi} \,f(z_0) \,d\theta$$

$$if(z_0) \oint_{0}^{2\pi} d\theta=2\pi i f(z_0)\text{ W.R.T}$$