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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 $t$.

Thanks

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"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 $d_\mathbb{H} (z_1,z_2) = \int_\gamma \frac{2i}{z - \bar{z}} \sqrt{dz d\bar{z}}$ –  user38268 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. –  user38268 May 13 '11 at 23:43

1 Answer 1

up vote 8 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 $\sqrt{(\frac{dx}{dt})^2 + (\frac{dy}{dt})^2}~dt$

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