On a property of holomorphic functions

We can see that a holomorphic functions are the complex functions such that they can express a convergent power series

\displaystyle f(z) = \sum_{i=0}^\infty a_nz^n.

We want to compute {\int\limits_\gamma f(z)dz}. Suppose {\gamma} is closed curve (Jordan curve).

We compute the integral for {\gamma = \mathbb{S}^1}. Firstly, consider {\int_{\mathbb{S}^1}z^ndz, n \ge 1}. Change the variables, {z = e^{2\pi it}, t \in [0,1]}, so

\displaystyle \int_{\mathbb{S}^1}z^ndz = \int_0^1e^{2n\pi it}2 \pi i\cdot e^{2\pi it}dt = \int_0^1e^{2(n+1)\pi it}2 \pi idt = \dfrac{e^{2(n+1)\pi it}}{n+1}\bigg|_0^1 = 0.

This computation, we think that it is important, because it implies a classical result
Theorem 1 (Cauchy integral theorem) Let {f : \mathbb{C} \rightarrow \mathbb{C}} be a holomorphic function on {\mathbb{C}}. Then

\displaystyle \int_\gamma f(z) dz= 0,

with {\gamma} is closed Jordan curve.

Of course it has modulo an any closed Jordan curve: If \gamma_1 and \gamma_2 are two closed Jordan curve with a common fixed point then  \displaystyle \int_{\gamma_1} f(z)dz = \int_{\gamma_2} f(z) dz,


Leave a Reply

Fill in your details below or click an icon to log in:

WordPress.com Logo

You are commenting using your WordPress.com account. Log Out / Change )

Twitter picture

You are commenting using your Twitter account. Log Out / Change )

Facebook photo

You are commenting using your Facebook account. Log Out / Change )

Google+ photo

You are commenting using your Google+ account. Log Out / Change )

Connecting to %s