Sign up ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

My book says that if $f(z)$ is analytic in a simply connected domain, then integration depends only on the endpoints, not on the path.

But I'm confused. I'm trying to imagine $f(z)$ as a function in the $x,y,z$ plane where $x$ and $y$ (in $z=x+yi$) correspond to the $x$ and $y$ axes, and $f$ corresponds to the $z$-axis. In the function $f(z)=1$, no matter how I imagine it, the area under the path is always larger when the path of integration is longer. What am I doing wrong?

share|cite|improve this question
Thanks for the answers. All of them were extremely helpful, and helped me to finally understand. My problem was that I wasn't taking into account orientation. – john smith Dec 29 '11 at 3:53

6 Answers 6

up vote 2 down vote accepted

To interpret the integral as the area under the "curtain", as you have, we must have $f:\mathbb C\to \mathbb R$ (rather than $f:\mathbb C\to\mathbb C$). Assuming this to be the case to find the area under the "curtain" we must compute the "line integral of a scalar field".

That is we consider $\mathbb C$ as vectors in $\mathbb R^2$, so that $f:\mathbb R^2\to\mathbb R$. Then we have:

$$\int_\gamma f(z) dz = \int_a^b f(\gamma(t))|\gamma'(t)|dt$$

where the contour is $\gamma:[a,b]\to\mathbb R$.

However, if we had instead computed a "complex line integral" we would have:

$$\int_\gamma f(z) dz = \int_a^b f(\gamma(t))\gamma'(t)dt$$

Notice there is no absolute value around the $\gamma'(t)$ term.

Considering the case where $f(z)=1$ the line integral of a scalar field becomes: $$\int_a^b |\gamma'(t)|dt$$

while the complex line integral becomes: $$\int_a^b \gamma'(t)dt$$

Since $|\gamma'(t)|\geq 0$ the area under the first integral will "increase for longer paths", the second integral will not. (Speaking informally).

share|cite|improve this answer

You are thinking of the wrong type of integration. Actually it is unclear what precisely you are thinking of, but it is not the correct type of integration.

First, remember that $f$ has complex values as well as complex input. When we write "$f(z)$", "$z$" is the input, or argument, what goes into the function, while the whole term $f(z)$ refers to the value of the function, or the output. You could also write $w=f(z)$ to make explicit that $z$ and $w$ are respectively the input and output variables, and $f$ sends the complex number $z$ to the complex number $w$.

But even in the real-valued case, a line integral along a curve is not typically interpreted as an area. In the complex-valued case, "area under the path" makes even less sense. The integral of $f$ along a path $\gamma$ can be defined as a Riemann-Stieltjes integral. In the case where $\gamma:[a,b]\to\mathbb C$ is a piecewise continuously differentiable path, it turns out that $\int_{\gamma}f(z)dz=\int_a^b f(\gamma(t))\gamma'(t)dt$. For example if $\gamma:[0,2\pi]\to\mathbb C$ parametrizes the unit circle with $\gamma(t)=e^{it}$, and if $f(z)=1$, then $\int_\gamma f(z)dz=\int_0^{2\pi}ie^{it}dt=0$.

Your book is correct, and this is a consequence of Cauchy's theorem. If you have $2$ different paths $\gamma_1$ and $\gamma_2$, that start and end at the same place, then if you take $\gamma_1$ followed by the reverse of $\gamma_2$, you get a closed curve in your domain, and the integral of $f$ on the closed curve is $0$ by Cauchy's theorem. If you unwind this it shows that $\int_{\gamma_1}f(z)dz=\int_{\gamma_2}f(z)dz$.

share|cite|improve this answer

Integration in the complex plane unfortunately doesn't correspond to the area under a curve. Instead, one physical interpretation for it is something like the work done to move along a curve.

Think about electric fields and electric potential. In this interpretation, the electric field takes the place of the function you want to integrate, and the potential takes the place of the result you get. Integration depends only on the endpoints, not the path itself: this basically says that no matter which way you move a charged particle through an electric field, you can calculate the total work done just by using the electric potential at the endpoints.

share|cite|improve this answer

If $z=x+iy$, you shouldn't be thinking of $f$ as corresponding to the $z$ axis. Both the argument to the function and the value of the function are in the $z$-plane.

If you had $\displaystyle \int 1\;|dz|$, you would just get the length of the path, and that of course depends on which path you follow, not just on the endpoints. But $\displaystyle \int 1\;dz$ is not "area under a curve", but rather it is just the net change as one moves along the curve.

share|cite|improve this answer

What you seem to be visualizing is the arc length of the path, which of course does not depend only on the endpoints. What's going wrong is that a path integral in complex analysis does not compute the arc length.

If you are taking complex analysis, you have probably already seen things like Stokes/Green's Theorem from multivariable calculus. There, you saw a similar phenomenon - indeed, what is going here is nothing but a special case. So you might think back to Stokes' Theorem for a more reliable bit of intuition.

share|cite|improve this answer

The error is in forgetting the $dz$ term in $\int f(z)\ dz$
Since $f(z)=1$ we are adding $\epsilon\ e^{i \theta} $ terms. You may imagine this as adding tiny vectors in a plane : the sum of these vectors is simply the vector starting from the tail of the first vector and finishing at the arrow of the last one.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.