Line integration in complex analysis

Question

In normal line integration, from what I understand, you are measuring the area underneath $f(x,y)$ along a curve in the $x\text{-}y$ plane from point $a$ to point $b$.

But what is being measured with complex line integration, when you go from a point $z_1$ to a point $z_2$ in the complex plane?

With regular line integration I can see $f(x,y)$ maps $(x,y)$ to a point on the $z$ axis directly above above/below $(x,y)$.

But in the complex case, when you map from the domain $Z$ to the image $W$, you are mapping from $\mathbb{R^2}$ to $\mathbb{R^2}$ ...it is not mapping a point to 'directly above/below'...so I don't have any intuition of what is happening with complex line integration.

Answer 1

Forget about areas for the moment.

Consider the following situation: At the begin you are at the origin of the $x$-axis and have to compress a spring which is attached far away to the right. Assume that when the left end of the spring is at a given $x\geq0$ then it presses back with force $f(x)$. If the force were a constant $F$ then the work $W$ done when pushing a cart from $x_0$ to $x_1$ along the $x$-axis would be $W=F\cdot(x_1-x_0)$. But in our case the force is variable. When you compress the spring by pushing a cart to the right, and after some time you are at the point $a>0$ then the total amount $W$ of work done in this process is represented by $$W\doteq \sum_{k=1}^N f(\xi_k)\ (x_k-x_{k-1})\doteq \int_0^a f(x)\ dx\ ,$$ where $0=x_0 < x_1 < \ldots < x_N=a$ is a partition of the interval $[0,a]$, and $x_{k-1}\leq\xi_k\leq x_k$ $\ (1\leq k\leq N)$.

Analogously in the complex domain for the purpose of line integrals you should not consider the given $z\mapsto f(z)$ as a mapping of the $z$-plane to some other domain, but as a "complex scalar field" which defines at each point $z\in{\rm dom}(f)$ a certain "complex force" $f(z)$. For a constant such force $F\in{\mathbb C}$ the "complex work" done when pushing a cart from $z_0$ to $z_1$ along a straight line is given by $F\cdot(z_1-z_0)\in{\mathbb C}$, where $\cdot$ denotes the ordinary product in ${\mathbb C}$.

Assume now that you are given a curve $$\gamma:\quad t\mapsto z(t)\qquad(a\leq t\leq b)\ .$$ Then the total "complex work" done when you push a cart along this curve would be represented by $$W\doteq \sum_{k=1}^N f\bigl(z(\tau_k)\bigr)\bigl(z(t_k)-z(t_{k-1})\bigr)\doteq \int_a^b f\bigl(z(t)\bigr) z'(t)\ dt =:\int_\gamma f(z)\ dz\ .$$

Answer 2

Think of it as separately tallying the real and imaginary parts of the function. That is, the path integral

$$ S=\int_C f(z) dz $$ could be written as a sum $S=R+iI$ of the real and imaginary parts $$ R=\int_C \Re[f(z)] dz $$ and $$ I=\int_C \Im[f(z)] dz $$

This works for $n$-dimensional coordinate integration as well, where you are separately tallying each coordinate of the vector-valued function $f$, though by the time you get there you are well on your way to connections on manifolds and the cliffs of insanity.

Answer 3

Adapted from page-481 of Tristan Needham's Visual Complex Analysis

The easiest way to understand the Complex Integral is through something known as Poly Vector fields. The essential idea can be split into three parts:

1. A complex function can be thought of as both as a point and a complex vector. We work with the second interpretation for understanding Complex integrals here. Visually, the situation looks like attaching an arrow at each point on the curve we integrate over.

2. For any complex function $H \equiv r e^{i \beta}$, we can define a Polya vector field of the form $\overline{H} \equiv |H| e^{-i\beta}$

3. After this, the the complex integral can be simplified into a more revealing form:

$$ \int H dz = \int |H| e^{i\beta} ds e^{i \alpha}= \int |H| ds e^{ i \left[ \alpha - (- \beta) \right]}= \int |H| ds e^{i \theta}= \int |H| ds \cos \theta+ i|H| ds \sin \theta $$

The first term on the right most equaliy can be thought of as the dot product of the complex vector $H$ vector with the tangent of the curve, and, the second with the unit normal. We have:

$$ \int Hdz = \int ds \left[ (H \cdot T) +i(H \cdot N)\right]$$

We may recall from Multivariable calculus that the first term is work along the contour and the second term is the flux.