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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.

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It is not just that the function is not mapped "above" or "below", that actually doesn't change the intuition much (you can treat the real and imaginary parts separately if that were the only difference). The main difference is that the line elements $ds$ and $dz$ for the two integrations are very different. –  Willie Wong Feb 17 '12 at 14:26
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Instead of comparing with the scalar line integral, a much closer relative of the complex line integral is line integrals for vector fields. –  Willie Wong Feb 17 '12 at 14:28

2 Answers 2

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 $$W\doteq \sum_{k=1}^N f\bigl(z(\tau_k)\bigr)\bigl(z(t_k)-z(t_{k-1})\bigr)\doteq \int_a^b f\bigl(z(t)\bigr) z'(t)\ dt =:\int_\gamma f(z)\ dz\ .$$

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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.

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