Proof of Cauchy Integral Formula

[alyafey22]

I want to prove the Cauchy integral formula :

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

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

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

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

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

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

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

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

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

 

[jvicens]

z_0Hello! Thank you for your interest in proving the Cauchy integral formula. I am always excited to see individuals taking an interest in understanding and proving mathematical concepts.

The Cauchy integral formula is an important result in complex analysis that relates the values of a function inside a closed curve to its values on the curve itself. It is often used in the evaluation of complex integrals and has many applications in physics and engineering.

In order to prove the Cauchy integral formula, we will use the concept of residues. The residue of a function at a point is defined as the coefficient of the term with a negative power in the Laurent series expansion of the function around that point. In simpler terms, it is the coefficient of the term with a negative power in the Taylor series expansion of the function around that point.

Now, let's consider the function f(z) and a closed curve \gamma that contains the point z_0. According to the Cauchy integral theorem, we have:

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

where \text{Res}(f(z),z_0) is the residue of the function f(z) at the point z_0.

Since our goal is to prove the Cauchy integral formula, we will take a closer look at the residue of the function f(z) at the point z_0. As mentioned earlier, the residue is the coefficient of the term with a negative power in the Taylor series expansion of the function around that point. So, we can write the Taylor series expansion of f(z) around z_0 as:

f(z)=f(z_0)+f'(z_0)(z-z_0)+\frac{f''(z_0)}{2!}(z-z_0)^2+...

The coefficient of (z-z_0)^{-1} in this expansion is given by \frac{f'(z_0)}{1!}. Therefore, the residue of f(z) at z_0 is \text{Res}(f(z),z_0)=f'(z_0).

Now, using the Cauchy integral theorem, we have:

\oint_{\gamma} \, \frac{f(z)}{z-z_0}\,dz