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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    $\begingroup$ 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. $\endgroup$ Feb 17, 2012 at 14:26
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    $\begingroup$ Instead of comparing with the scalar line integral, a much closer relative of the complex line integral is line integrals for vector fields. $\endgroup$ Feb 17, 2012 at 14:28

4 Answers 4


The most important calculation in all of complex analysis is the following:

$$ \int_C \frac{1}{z}dz = 2\pi i $$

Where $C$ is the unit circle centered at the origin. Let's see if we can understand this integral geometrically. Hopefully this will illuminate the general meaning of the complex integral as well.

Cut the unit circle up into a million tiny directed line segments (all pointed in the "positive" direction). Each one of these is a small change in $z$ along the curve: each one of these is a little "$dz$". The directed line segment from the origin to that point on the unit circle is $z$. For points on the unit circle, $\frac{1}{z} = \overline{z}$ is the reflection of $z$ in the real axis. Remember that complex numbers multiply by multiplying the moduli and summing the arguments. Since if the argument of $z$ is $\theta$, then the argument of $dz$ is $\frac{\pi}{2}+\theta$ by geometry (radius is perpendicular to tangent of a circle), and the argument of $\frac{1}{z}$ is $-\theta$. So the argument of $\frac{1}{z}dz$ is $\frac{\pi}{2}$! Also since $|z|=1$, $|\frac{1}{z}dz|=|dz| = ds$, the length of the little line segment. Thus $\frac{1}{z}dz$ is a little line segment of length $ds$ pointing straight up, where $ds$ is the length of a little arc of the circle. In other words, $\frac{1}{z}dz = i ds$ Summing all of these around the unit circle we get $2\pi i$.

Hopefully this picture helps you to visualize all of this:

enter image description here

  • $\begingroup$ I don't see how this helps. This just describes in word the standard parametric calculation of the value of the integral, not what it represents. $\endgroup$ Feb 29, 2020 at 0:06

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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    $\begingroup$ As far as I can see, all this answer does is explain the definition of complex integration, and attach the word "complex work" to it. Despite being kind of catchy, I do not see how this conveys any more intuition. How does it relate to "real work"? $\endgroup$ Aug 31, 2014 at 20:46

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.


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 $\overline{H} \equiv |H| e^{-i\beta}$ enter image description here

  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.


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