Explanation of complex integration in plain english?

mathperson

Im trying to get my head around complex integration/complex line integrals. Real integration can be thought of as the area under a curve or the opposite of differentiation. Thinking of it geometrically as the area under a curve or the volume under a surface in 3 dimension is very intuitive.

1. So is there a geometric way of thinking about complex integration? Or should I just be viewing it as process that reverses differentiation? Or has integration other meanings in complex analysis?


2. Here is an example, could someone explain this to me -
Here's the definition of the integral along a curve gamma in C, parameterized by w:[a, b] ->C
[latex]\int_\gamma f(z)dz = \int^b_a f(w(t)).w{(t)}'dt[/latex]

So I have -
[latex]\gamma[/latex] is the unit circle with anti-clockwise orientation parameterized by [latex]w:[0, 2\pi]\to C[/latex]
[latex]w(t) = e ^{it} = Cos(t) + iSin(t)[/latex]

So if use the definition of the integral, [latex]\int_\gamma f(z)dz = \int^b_a f(w(t)).w{(t)}'dt[/latex], and work this out it comes to

[latex]\int^{2\pi}_0 i dt = 2{\pi}i[/latex]

So what does this [latex]2{\pi}i[/latex] represent? Does it mean anything geometrically, like if a regular integral works out to be 10 that means the area under the curve between 2 points is 10...

Michael Collins

You can always write

[latex] \displaystyle \int_{\gamma}f(z)\,\hbox{d}z = \int_{\gamma}\left[u(x,y)+\hbox{i}v(x,y)\right]\, \hbox{d}(x+\hbox{i}y) = \int_{\gamma}\left[u(x,y)\,\hbox{d}x-v(x,y)\,\hbox{d}y\right] + \hbox{i}\int_{\gamma}\left[u(x,y)\,\hbox{d}x+v(x,y)\,\hbox{d}y\right] [/latex]

which are scalar vector line integrals, with the usual geometric meaning.

We can go further though, looking at just the imaginary part above

[latex] \int_{\gamma} \displaystyle \left[u(x,y)\,\hbox{d}x+v(x,y)\,\hbox{d}y\right] = \int_{\gamma} \left[u(x,y)\,{\bf\hat{x}}+v(x,y)\,{\bf\hat{y}}\right]\cdot \left(\hbox{d}x\,{\bf\hat{x}} + \hbox{d}y\,{\bf\hat{y}} \right) = \int_{\gamma} \left[u(x,y)\,{\bf\hat{x}}+v(x,y)\,{\bf\hat{y}}\right]\cdot \hbox{d}{\bf \hat{r}} [/latex]

So this means the imaginary part of the complex integral of the function [latex] f(z) = u+\hbox{i}v [/latex] is equavlent to a vector line integral of the vector valued function

[latex]{\bf\hat{F}} = u\,{\bf \hat{x}}+v\,{\bf \hat{y}} [/latex]

Clearly the real part corresponds to something similar - actually it is a vector field identified with the complex conjugate of the function [latex] f(z) [/latex].