Calc 2 babby here. Can someone explain what does this mean?

>>8635534
https://en.wikipedia.org/wiki/Line_integral

The thing about complex analysis, is that you're integrating over a 4-dimensional space, or basically, something that resembles R^4.

This means there is no intuitive way of imagining complex integration, other than thinking of how line integrals work in R^3.

Imagine you're in 3D space, and draw a curve on it. Then when you integrate, you're finding the area of the lets say, Riemann rectangles under the curve. Now imagine this curve is a closed loop. You're still finding the area under the curve.

You can draw a curve in any dimensional space with 2<=n, so you can define line integrals for any such dimensional space, and in particular, n=4 (with n=2 being your normal babby integral)

>>8635586
Not really. You're integrating over a 2D space with a 2D vector field defined on it. It's really just like Green's theorem.

When you do a closed loop integral that evaluates to 0 what it means is that you're just doing a line integral in a conservative field.

Complex analysis is mostly just 2D vector calculus.

>>8635586
I think I kinda understand. The analogy with 3D space makes it a bit understandable but I am still lacking intuition. I mean, when I set up integrals for normal calculus problems I rely a lot on intuition. For example, If I want to find the volume of revolution of a curve I do a rough sketch and do a lot of labelings on my graph to know how to set up the integral in a way that it will give me the actual volume I want. How can I bring this intuition for complex integrals.

A simpler example I was reading is how if you define the function
[eqn] f(z) = \frac{1}{z} [/eqn] and integrate it over the circle [math] |z| = 1 [/math] the answer you get is [math] 2 \pi i [/math].

What is the intuition behind that? From what I see is that if I grab that function over that circle, what I get is the same circle. The function just shuffles the points around.

Then I also noticed that the result of the integral of that function over that circle is the circumference of that circle, multiplied by i.

Does this mean anything? Is it just a coincidence? What is the intuition behind this result?

>>8635592
Pi can never be a perfect circle

Phi on the other hand..

>>8635591
you're taking a function of two variables, which has an output of two variables. To be able to graph the function, you must have 4 axes.