# Geometric interpretation of complex path integral

Let's say that we want to make sense of integrating a function $f: \mathbb{C}\rightarrow\mathbb{C}$ over some path $\gamma$. I can imagine two reasonable ways of doing it. First, there's the way that we all know and love: $$\int_\gamma f := \int_a^b f(\gamma(t))\gamma'(t)\,dt$$ where $\gamma(t)$ parametrizes $\gamma$.

Alternatively, we could say, "We already have a nice formulation of line integrals for paths in $\mathbb{R}^n$, and $\mathbb{C}$ is essentially just $\mathbb{R}^2$." So we decide that we'll define the path integral like this: $$\int_\gamma f := \int_a^b f(\gamma(t))|\gamma'(t)| \,dt$$ like we might do for functions $f: \mathbb{R}^2 \rightarrow \mathbb{R}$. This second definition has the nice property that $\int_\gamma f = \int_\gamma \text{Re}f + i \int_\gamma \text{Im}f.$

But of course we don't use the second formulation. We stick with the first, and there are some nice utilitarian reasons for doing so (Cauchy's integral formula, etc.). But are there more natural (geometric, preferably) reasons for using the first definition above?

EXAMPLE

For a simple example, let's say that $f(z) = i$ and I want to integrate $f$ along $\gamma$, the line from $0$ to $i$. I'll choose the parametrization $\gamma(t) = it$, $t \in [0,1]$. Then $\gamma'(t) = i$, and, using the first definition for our path integral, we get: $$\int_\gamma f = \int_0^1 i \gamma'(t) \,dt = \int_0^1 i^2\, dt = -1.$$

Meanwhile, in the wayward world where we choose the second formulation for our path integrals, we have: $$\int_\gamma f = \int_0^1 i |\gamma'(t)| \,dt = i.$$ In essence, this second computation didn't care about the complex structure on $f$'s domain. As far as this method was concerned, we integrated a constant function over a path of measure 1, so naturally we get that constant as our answer. On the other hand, in the first computation, multiplying by $\gamma'$ carries some additional complex structure. It knows the difference between integrating along lines from $0$ to $i$ or from $0$ to $1$, for example.

Heuristically, it seems like the first definition allows the complex structures on $f$'s domain and range to interact (through the $\gamma'$ term), whereas the second definition neglects any complex structure on the domain of $f$ and is only concerned with stretching/reorientation caused by our choice of parametrization. But why, a priori, are we interested in allowing $f$'s complex domain and range to interact in this way (if that's even the right way of framing it)? Does this "interaction" have a geometric interpretation? And is there a nice way of interpreting the line integral that makes the first definition obviously the more natural choice, or do we just go with it because it's useful?

Thanks for reading!

• As you say, the second definition ignores the fact that the domain is $\mathbb{C}$. So it is just a path integral in a two dimensional space, sometimes called a line integral. – Matt Aug 20 '14 at 21:50
• The first one tells us that if we have a square of "height" $i$ and "width" $i$ then the "area" is -1. It seems to me to be more the way that complex numbers should interact. – Paul Sundheim Aug 20 '14 at 21:52
• As we all know, sometimes the best answer to a question is a link to a text that has already been written by someone else, and which is very clear and accurate. In this case, I would refer to this very page! The so-called question includes all the explanations and answers, and is much fun to read. I learnt a lot from it. Cheers! – Amitai Yuval Aug 20 '14 at 22:03
• Nice question, +1. – BW. Aug 20 '14 at 22:41
• Hi Josh, just a comment to point out that I have added to my answer. – Mike F Sep 10 '14 at 20:30

## 2 Answers

I thought about this once and came up with an "answer" of sorts.

Notation: Given $z=a+ib$ and $w = c+id$ in $\mathbb{C}$, I will write $z \cdot w$ for the Euclidean dot product $ac + bd$. Ordinary multiplication of complex numbers will be denoted by juxtaposition.

The basic observation is, given $z,w \in \mathbb{C}$, the quantity $\overline z w$ is related to the dot product as follows: \begin{align*} \mathrm{Re}(\overline z w) = z \cdot w && \mathrm{Im}(\overline z w) = (iz) \cdot w. \end{align*} Note that $iz$, geometrically speaking, is $z$ rotated counterclockwise by 90 degrees.

Now, rather than interpret $\int_\gamma f$, let us interpret $I = \int_\gamma \overline f$. As you noted, $$I= \int_0^1 \overline{f(\gamma(t))} \gamma'(t) \ dt,$$ and so \begin{align*} \mathrm{Re}(I) = \int_0^1 f(\gamma(t)) \cdot \gamma'(t) \ dt && \mathrm{Im}(I) = \int_0^1 (if(\gamma(t))) \cdot \gamma'(t) \ dt \end{align*} or, in the notation of line integrals of vector fields, \begin{align*} \mathrm{Re}(I) = \int_\gamma f(z) \cdot dz && \mathrm{Im}(I) = \int_\gamma (if(z)) \cdot dz. \end{align*} So, for instance, thinking of $f$ as a force field,

• the real part of $\int_\gamma \overline f$ is the work done traveling along $\gamma$ through the force field $f$,
• the imaginary part of $\int_\gamma \overline f$ is the work done traveling along $\gamma$ through the force field $if$.

Note $if$ is just the force field $f$ rotated counterclockwise by 90 degrees.

Hope this helps...

Added: I thought I would add a simple example.

Put $f(z) = \overline{1/z}$. As you are no doubt aware, if the contour $\gamma$ is closed and avoids the origin, then $\int_\gamma \overline f = \int_\gamma \frac{1}{z}$ is equal to $2 \pi i$ times the winding number of $\gamma$.

In this example, the vector field $f(z) = \overline{1/z}$ is the gradient of the scalar potential $V(x,y) = \frac{1}{2} \log (x^2+y^2)$, so no work is done going around a contour in this field. The field $f$ looks like this: On the other hand, the 90 degree rotated field $if$ looks like this: Added: I just learned the proper name for this interpretation. The complex conjugate $\overline f$, viewed as a vector field, is called the "Polya vector field of $f$". As I mentioned, the real part of $\int_\gamma f$ is the work done by the Polya vector vield on a particle as it travels along the curve $\gamma$. An alternative way to think of the imaginary part of $\int_\gamma f$ is as the flux of the Polya vector field through the oriented curve $\gamma$. These ideas are explained in Chapter 11 of Tristan Needham's book "Visual Complex Analysis".

• Thank you for taking the time to put this together! Your interpretation and example helped a bunch. – Josh Keneda Aug 21 '14 at 20:07

As I commented a few days ago, I think that many answers to this question were already given by the person who asked it. However, it occurred to me that there is one very simple fact which may be relevant to this discussion, and wasn't mentioned yet.

As we all know, if $F,f:\mathbb{R}\to\mathbb{R}$ are two real functions such that $F'=f$, then $$\int_a^bfdx=F(b)-F(a).$$ This is extended, of course, to functions from a higher dimensional Euclidean space to $\mathbb{R}$ by using the gradient: $$\int_\gamma\nabla f\cdot d\gamma=f(\gamma(b))-f(\gamma(a)),$$ or alternatively, without using any metric structure,$$\int_\gamma df(\partial/\partial\gamma)=f(\gamma(b))-f(\gamma(a)),$$where $f$ is $\mathbb{R}^n\to\mathbb{R}$ and $\gamma$ is a differentiable path.

Similarly, if $F:\mathbb{C}\to\mathbb{C}$ is holomorphic with derivative $f$, and $\gamma:[a,b]:\to\mathbb{C}$, we have $$\int_\gamma fd\gamma=F(\gamma(b))-F(\gamma(a)).$$ The other definition (which ignores the complex structure of the domain, as has been said) does not satisfy this last equation, and in this manner lacks an essential property of integration.

• Good point! Thanks for commenting. – Josh Keneda Aug 25 '14 at 23:38