Independence of Path

**UlyssesMac** · November 5th 2010, 07:06 AM

i am a little confused by the idea of "independence of path" in complex analysis. i understand that certain functions, when integrated over simple paths with the same endpoints a and b yield the same value. however i do not understand how to determine whether the function is independent of the path (without calculating over arbitrary simple paths).

can anyone help enlighten me? thanks in advance!

**chisigma** · November 5th 2010, 07:31 AM

A fundamental theorem due to the French mathematician Augustin Cauchy extablishes that if $f(z)$ is analytic in a region inside a closed path $\gamma$ is...

$\displaystyle \int_{\gamma} f(z)\ dz =0$ (1)

Now if $f(z)$ is analytic in the whole complex plane [i.e. is an entire function...] the (1) is valid independently from $\gamma$ , and that means that, if we devide $\gamma$ in two parts, one coming from $z_{1}$ to $z_{2}$ and the other coming back from $z_{2}$ to $z_{1}$ is...

$\displaystyle \int_{z_{1}}^{z_{2}} f(z)\ dz + \int_{z_{2}}^{z_{1}} f(z)\ dz =0$ (2)

... and that means that $\displaystyle \int_{z_{1}}^{z_{2}} f(z)\ dz$ depends only from $z_{1}$ and $z_{2}$ and doesn't depend from the path connecting $z_{1}$ and $z_{2}$ ...

Kind regards

$\chi$ $\sigma$

**Drexel28** · November 5th 2010, 07:36 AM

Originally Posted by chisigma

A fundamental theorem due to the French mathematician Augustin Cauchy extablishes that if $f(z)$ is analytic in a region inside a closed path $\gamma$ is...

$\displaystyle \int_{\gamma} f(z)\ dz =0$ (1)

Now if $f(z)$ is analytic in the whole complex plane [i.e. is an entire function...] the (1) is valid independently from $\gamma$ , and that means that, if we devide $\gamma$ in two parts, one coming from $z_{1}$ to $z_{2}$ and the other coming back from $z_{2}$ to $z_{1}$ is...

$\displaystyle \int_{z_{1}}^{z_{2}} f(z)\ dz + \int_{z_{2}}^{z_{1}} f(z)\ dz =0$ (2)

... and that means that $\displaystyle \int_{z_{1}}^{z_{2}} f(z)\ dz$ depends only from $z_{1}$ and $z_{2}$ and doesn't depend from the path connecting $z_{1}$ and $z_{2}$ ...

Kind regards

$\chi$ $\sigma$

It doesn't need to be entire. Only holomorphic on a path connected subspace.

**chisigma** · November 5th 2010, 08:19 AM

Originally Posted by Drexel28

It doesn't need to be entire. Only holomorphic on a path connected subspace.

In order to avoid misunderstanding a simple example. Susppose that is $f(z)=\frac{1}{z}$ , $z_{1}= -i\ \sqrt{2}$ , $z_{2}= i\ \sqrt{2}$ and $\gamma$ is the 'red circle' in the figure...

By the residue theorem is...

$\displaystyle \int_{\gamma} \frac{dz}{z} = 2\ \pi\ i$ (1)

... and that means that $\displaystyle \int_{z_{1}}^{z_{2}} \frac{dz}{z}$ will be $\pi i$ if the point $z=0$ is on the left of the half circle connecting $z_{1}$ to $z_{2}$ and $- \pi\ i$ if it is on the right...

Kind regards

$\chi$ $\sigma$

**roninpro** · November 5th 2010, 08:43 AM

I think that Drexel is partially correct. You don't need the function to be entire. However, it must be defined on an open simply-connected domain.

See Cauchy's integral theorem - Wikipedia, the free encyclopedia.

**Drexel28** · November 5th 2010, 08:55 AM

Originally Posted by roninpro

I think that Drexel is partially correct. You don't need the function to be entire. However, it must be defined on an open simply-connected domain.

See Cauchy's integral theorem - Wikipedia, the free encyclopedia.

Ah crap. I always say the wrong hypothesis. You'd think someone who's interested in alg. top. I'd remember haha

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