- 21 Apr '10 10:45 / 2 editsLets say f(x) = 1. If we integrate f over some finite domain, we get the length of that domain. Lets extend this to the complex plane and say f(z) = 1. If we integrate f around the unit circle centered at the intersection of the real and imaginary axes we get 0. I do not understand the meaning of this integral. I would have expected to get the circumference of the circle: 2*pi. What is going on here?

Is it because when multiplying f(z)dz dz is negative in the real and imaginary parts at just the right places to give 0? - 21 Apr '10 11:38 / 1 editYou are calculating the residual of the function f(z) = 1. This function has no poles and the residual is zero.

The real analog of this intergration would be to integrate f(x) = 1 from -1 to 1, and then from 1 back to -1 (obviously giving zero). Maybe this analog is a bit handwaving, but just to get you a feel of what you're doing.