Cauchy integral theorem

1. May 20, 2008

antonantal

I was looking at the proof of the residue theorem on MathWorld: http://mathworld.wolfram.com/ResidueTheorem.html
and got stucked on the 3rd relation.

I don't understand why the Cauchy integral theorem requires that the first term vanishes.
From the Cauchy integral theorem, the contour integral along any path not enclosing a pole is 0. But in this case, the contour $$\gamma$$ encloses $$z_{0}$$ which is a pole of $$(z-z_{0})^{n}$$ for $$n \in \{-\infty,...,-2\}$$

2. May 20, 2008

nicksauce

Perhaps they meant the Cauchy Integral Formula

3. May 20, 2008

Crosson

I agree with nicksauce. To expand his point go to this page and scroll down to equation 19 at the bottom:

http://mathworld.wolfram.com/CauchyIntegralFormula.html

In this case $f(z) = 1$ for all z, and so the sum of these terms vanishes because the derivatives of a constant are all zero.

Update: I sent a message to the mathworld team suggesting the following correction:

Below equation (3) the given justification: "The Cauchy integral theorem requires that the first and last terms vanish" should be replaced with something to the effect of "The first term is zero because of the Cauchy Integral Formula, while the last term is zero because of the Cauchy Integral Theorem." Thank you for hosting the only elementary proof of even this much of the residue theorem that I could find anywhere on the web.

Last edited: May 20, 2008
4. May 20, 2008

antonantal

Thanks! Good idea to send that suggestion to the MathWorld team too.