What is a geometric explanation of complex integration in plain English?

Question

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 $w:[a, b] ->C$ \begin{equation} \int_\gamma f(z)dz = \int_a^b f(w(t)).w{(t)}'dt \end{equation}

So I have -

$\gamma$ is the unit circle with anti-clockwise orientation parameterized by $w:[0, 2\pi]\to C$

$w(t) = e ^{it} = Cos(t) + iSin(t)$

So if use the definition of the integral,

$\int_\gamma f(z)dz = \int^b_a f(w(t)).w{(t)}'dt$, and work this out it comes to

$\int^{2\pi}_0 i dt = 2{\pi}i$

So what does this $2{\pi}i$ 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...

Answer 1

In every book or course where line integrals occur (in physics, complex analysis, geometry, $\ldots$) people try to explain in so many (english!) words what a line integral is. It's not always the same thing and comes in various forms. As a rule of thumb one can say the following:

A line integral is a function that assigns to any (real scalar, complex scalar, vector) field $f$ on a domain $\Omega$ and any (directed) curve $\gamma\subset\Omega$ a certain value $v$, denoted by $$\int_\gamma f(x)* dx$$ (or similar). This rule should have the following properties; the first one giving the geometric or physical intuition behind $v$:

1. When $f$ is constant and $\gamma$ is the segment with initial point $x_0$ and endpoint $x_1$ then $v=f*(x_1-x_0)$.

2. The value $v$ is independent of the chosen parametrization of $\gamma$.

3. When $\gamma=\gamma_1+\gamma_2$ in an obvious way then $$\int_\gamma f(x)* dx =\int_{\gamma_1} f(x)* dx +\int_{\gamma_2} f(x)* dx\ .$$

4. When $f$ and $g$ are two such fields then $$\int (f+g)*dx=\int_\gamma f*dx +\int_\gamma g*dx\ ,\qquad \int_\gamma (\lambda f)* dx=\lambda\int_\gamma f*dx\ .$$

Here $*$ denotes any multiplication that makes sense in the actual situation, and $dx$ might as well be $|dx|$ in certain cases.

Answer 2

A good intuition could be to think of an integral as some kind of "mean value" instead of a volume: The value $\int_0^1 f(x)dx$ is the "mean value of the function $f$ over the interval $[0,1]$". (Resorting to area again: For non-negative $f$ the area under the graph of $f$ is equal to the rectangle with width 1 and height $\int_0^1 f(x)dx$.)

In a vague way, something similar holds for a complex line integral: The integral is "the mean value of $f$ along the curve $\gamma$, however, taking the infinitesimal directions of $\gamma$ into account".

Another way to view complex integrals is as integral of real 2-dimensional vector field (given by the complex valued function $f$) along paths in 2 dimensions (given by $\gamma$). The intuition is basically the same as "mean values" but here you could additionally think of "mean velocities" or "mean directions". Physically, the vector field describes a force, acting on a particle and the path describes the movement of the particle. The integral corresponds to the "mean forced which acted on the particle during motion".