Take the 2-minute tour ×
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.

Just a quick question, I am going through some books on complex analysis and I'm wondering what an integral like $\int_f f(z,\bar{z}) \sqrt{dz d\bar{z}}$ means. How is one supposed to take that notation to mean something?? I have not come across things that have roots over the $dz's$ and $d\bar{z}'s$. Besides, the integral here is a single integral, but there are two variables involved?

I am assuming I can do that as perhaps I can parametrise $f$ in terms of some parameter $t$.


share|improve this question
"going through some books on complex analysis" - which books? They ought to have explained their notation somewhere... –  J. M. May 7 '11 at 8:38
It looks like a plain-old contour integral to me, in awkward terminology. Could you supply a concrete example of a computation? –  Ryan Budney May 10 '11 at 22:28
@Hi just some corrections to the comment above. The first error should read $d_\mathbb{H} (z_1,z_2) = \int_\gamma \frac{2i}{z - \bar{z}} \sqrt{dz d\bar{z}}$ –  fpqc May 13 '11 at 23:42
What I meant to say was, "It is not that I do not know how to evaluate integrals, but the notation seems confusing and I don't get how one can evaluate some integral like this directly. Also, $\gamma$ is the semi circle passing through $z_1$ and $z_2$ whose center lies on the real - axis. –  fpqc May 13 '11 at 23:43
add comment

1 Answer

up vote 7 down vote accepted

This is somewhat of a guess... hopefully you can find the definition somewhere in the text. But I think that this notation means you are integrating with respect to path length. Write out the differentials in terms of real and imaginary parts, and $ \sqrt{dz ~ d\bar{z}} $ becomes $ \sqrt{dx^2 + dy^2} $. So if you plug $ f(z,\bar{z}) = 1 $ into your expression, this integral will give you the length of that path.

To make it totally concrete, if you parametrize your path with respect to some parameter $t$, then the differential becomes $\sqrt{(\frac{dx}{dt})^2 + (\frac{dy}{dt})^2}~dt$

share|improve this answer
add comment

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.