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Proof of Cauchy Integral Formula

ZaidAlyafey

Well-known member
MHB Math Helper
Jan 17, 2013
1,667
I want to prove the Cauchy integral formula :

[tex] \oint_{\gamma} \, \frac{f(z)}{z-z_0}\,dz= 2\pi i f(z_0) [/tex]

[tex]\text{so we will integrate along a circle that contains the pole .}[/tex]

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

[tex] z =z_0+\delta e^{i\theta }\,\, so \,\,dz =i\delta e^{i\theta }\,d\theta[/tex]

[tex]\text{We will make the radius as small as possible to contain the pole . }[/tex]

[tex]\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[/tex]

[tex]\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[/tex]

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


[tex]i \oint_{0}^{2\pi} \,f(z_0) \,d\theta[/tex]

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