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 $\mathbb{C}$, parameterized by $w(t):[a, b] \to \mathbb{C}$.

$$\int_\gamma f(z) \mathrm{d}z = \int_a^b f[w(t)] w'(t) \mathrm{d}t$$

So I have that:

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

  • $w(t) = e^{it} = \cos(t) + i \sin(t)$.

So if we use the definition of the integral,

$$\int_\gamma f(z) \mathrm{d}z = \int^b_a f[w(t)]w'(t) \mathrm{d}t$$

and work this out, it comes to

$$\int^{2\pi}_0 i \mathrm{d}t = 2{\pi}i$$

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...

  • $\begingroup$ This seems to be a duplicate of math.stackexchange.com/questions/110334/… $\endgroup$ Feb 20, 2012 at 20:45
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    $\begingroup$ I am asking for an explanation in 'plain english' (as much as possible) in this post as the answers in the other post were not intuitive to me. And specifically in this post I am asking what the result that I got - $2{\pi}i$ - actually means. If I got the derivative of a y = f(x) and it turned out to be 2 I know that that means y increase by 2 when x increases by 1. So what does the $2{\pi}i$ result that I got mean in the case of complex integration. $\endgroup$
    – Jim_CS
    Feb 20, 2012 at 21:06
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    $\begingroup$ I suggest that you simply stop worrying about this. Reason by analogy with what you know, and try to transfer tentatively your intuition about the real integral to the complex one—with some aspects you should have absolutely no trouble, and with others you will. When you hit one of this problematic pieces of intuition, then experience, practice and knowledge will construct something in your mind that you will only be able to describe as intuition.$$ $$Trying to wrap one's head around something one is not familiar enough and about which one is only beginning to learn is a very unnatural thing! $\endgroup$ Feb 21, 2012 at 8:14
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    $\begingroup$ Does this answer your question? Line integration in complex analysis $\endgroup$ Mar 16, 2022 at 17:12

5 Answers 5


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.


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".

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    $\begingroup$ Dear Dirk, I don't think so. If you take the path $\gamma(t)=it\; (0\leq t\leq 1)$ in $\mathbb C$ and $f(z)=1$, you get $\int_\gamma f(z)dz=\int_\gamma 1dz=i$. It is hard to believe that the mean value along the curve $\gamma$ of a function constantly equal to $1$ is $i$. $\endgroup$ Feb 20, 2012 at 20:39
  • $\begingroup$ Thanks, I forgot to take "directions" into account. Edited my answer... $\endgroup$
    – Dirk
    Feb 21, 2012 at 7:47
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    $\begingroup$ Note that $\int_0^1f(x)dx$ is only the mean value of of over the interval $[0,1]$ thanks to the fact that the interval has measure $1$. Otherwise, you'd have to divide your integral by the measure of the interval. $\endgroup$
    – user23211
    Feb 21, 2012 at 11:50

I know I am about a decade late to the party, but I am about to start teaching an undergraduate complex analysis course in a couple of days and have been trying to think about a geometric interpretation of the contour integral to help explain to people whose mathematical background is at the basic vector calculus level. Here is one thing I have come up with, that I think is more or less "dual" to the answer by @Dirk. I'll be curious to see if anyone finds it helpful.

If we write $z=x+iy$, then we can write a complex valued function as $f(z)=u(x,y)+iv(x,y)$. A slightly different point of view is that the function $f(z)$ defines a vector field on $\mathbb{R}^2$ given by $f(x,y)=\langle u(x,y),v(x,y)\rangle$. If $\gamma(t)=x(t)+iy(t)$ is a contour then by unpacking the definition of $\int_\gamma f(z)dz$ and breaking into real an imaginary parts we get a pair of line integrals:

$$\int_\gamma f(z)dz=\int_\gamma udx-vdy+i\left(\int_\gamma vdx+udy\right)=\int_\gamma \langle u,-v\rangle\cdot ds+i\int_\gamma \langle v,u\rangle \cdot ds,$$

where $\tilde ds=\langle dx,dy\rangle$ and $ds$ is the corresponding unit tangent vector to $\gamma$. In other words, $ds$ is the line element tangent to the curve $\gamma$. By a standard trick from vector calculus, we can rewrite a 2-dimensional line integral as a flux integral of a different vector field. More precisely, let $\tilde dn=\langle -dy,dx\rangle$ and $dn$ the corresponding unit vector. We can think of $dn$ as an outward unit normal element to the curve $\gamma$. A simple computation give $\langle u,-v\rangle\cdot ds=\langle v,u\rangle \cdot dn$, and so

$$\int_\gamma \langle u,-v\rangle\cdot ds=\int_\gamma \langle v,u\rangle\cdot dn.$$

Thus the real part of $\int_\gamma f(z)dz$ is the flux of the vector field $\langle v,u\rangle$ through the curve $\gamma$. Returning to the complex setting we can interpret the vector field $\langle v,u\rangle$ as the complex function $g(z)=i\overline{f(z)}$.

A similar computation for the imaginary part of $\int_\gamma f(z)dz$ shows that it is equal to the flux of the vector field $\langle -u,v\rangle$ through the curve $\gamma$. On the complex side we see that this vector field corresponds to the function $h(z)=-\overline{f(z)}$.

Note that $h(z)=-ig(z)$ and so these two function/vector fields just differ by rotation by $\pi/2$ With this perspective the contour integral $\int_\gamma f(z)dz$ comes from combining the fluxes of $g(z)$ and its rotation $h(z)$ through the curve $\gamma$.

If $f(z)$ is analytic this says that $\overline{f(z)}$ and the "rotated" function $i\overline{f(z)}$ both have zero flux through any closed curve.

If $f(z)=1/z$ then this says that $-\frac{1}{\bar{z}}$ thought of as a vector field on $\mathbb{R}^2$ has flux $2\pi$ through any curve that winds once around 0 and that the rotated vector field $\frac{i}{\bar{z}}$ has zero flux through any curve that winds once around 0.


The (right) intuition in Mathematics is not only an interesting effort but a necessary ability to perceive and create theory. Otherwise Mathematics seem like a vacuous game of symbols.

Let an integrable $f:U\to \Bbb C$, where $U\subset \Bbb C$ is open and contains a curve $γ:[a,b]\to \Bbb C$. If $dz$ is a vector "almost tangent" to $γ$ at $z\in γ([a,b])$, $\rm Arg$$f(z):=u$, $\rm Arg$$dz:=v$, $v-u:=θ$ the angle between $\overline {f(z)}$ and $dz$, then

$$f(z)dz=|f(z)||dz|(\cos(u+v)+i\sin (u+v))=$$ $$=|f(z)||dz|\cos(u+v)+i|f(z)||dz|\sin (u+v)=$$ $$=|\overline {f(z)}||dz|\cos(-u+v)+i|\overline {f(z)}||dz|\sin (-u+v)=$$ $$=|\overline {f(z)}||dz|\cosθ+i|\overline {f(z)}||dz|\sinθ=$$ $$=\overline {f(z)}\cdot dz + i(\overline {f(z)}\cdot (-idz)).$$

The real part of the above expression gives the element of work done by $\overline f$ along $dz$ and the imaginary part gives the flux element of $\overline f$ through $γ$ at $z$ (obviously $-idz$ can be represented as a vector normat to $γ$ at $z$). Integration along $γ$ gives

$$\int_γ f(z)dz=Work(\overline f, γ)+iFlux(\overline f, γ).$$

Of course this approach is a geometric one, for "work" is the total tangential vector field along the curve (for a closed curve we have the circulation) and "flux through" the curve is the total normal vector field through the curve.


Here is geometrical/kinematic model. It involves plotting curves in two distinct copies of the complex plane.

  1. Start with the original path $C$ of integration written in parametrized form as $C= z(t)$. This can be interpreted as the trajectory of a moving particle traveling in the $z$ plane. Then $\frac{dz}{dt}$can be regarded as the velocity vector in the complex plane.

  2. If your method for estimating speed and direction are inaccurate (due to a faulty clock and compass), then the incorrect velocity you might mistakenly record with your faulty instruments can be denoted as $\frac{dw}{dt}$. It is a locally dilated and possibly rotated version of the true velocity $\frac{dz}{dt}$. It can be checked that the local rotation and dilation can be accomplished by multiplying by some complex distortion factor $f(z)= A e^{i \theta}$. Here $A$ describes the dilation and $\theta$ describes the rotation. Thus we can write $\frac {dw}{dt}= f(z(t)) \frac{dz}{dt}$. ( This is what T. Needham has called the ampli-twist interpretation of multiplication by a complex factor in his book Visual Complex Analysis.)

  3. Reconstructing the imputed position vector in the $w$ plane (the path you would plot on a chart based on your faulty measurements) produces the new curve in the auxiliary $w$ plane defined by the indefinite integral $\tilde C(\tau)=w(\tau)=\int _0^\tau\frac{ dw}{dt} dt = \int f(z) \frac{ dz}{dt} dt =\int_C f(z) dz$.

Advanced Comment regarding Cauchy's Theorem. From this model one sees immediately that in general for a randomly selected non-analytic distortion function $f(z)$, the integral $\oint_C f(z) dz$ along a closed loop $C$ is generally not zero. Note that the condition that $\oint_C f(z)dz =0$ means that the net displacement in the $w$ plane vanishes when we traverse a closed loop $C$ in the $z$ plane. Cauchy's Theorem and its relation to the existence of a complex anti-derivative can be explained by interpreting the existence of a complex anti-derivative $w= F(z)$ as a geometric transformation of the $z$ plane to the $w$ plane that (if it exists) must map closed loops $C$ to closed loops $\tilde C$.


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