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Thread: Something I am having trouble understanding - complex analysis

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    MHF Contributor Bruno J.'s Avatar
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    Something I am having trouble understanding - complex analysis

    Consider the function $\displaystyle f(z)=\frac{1}{\sqrt{1-z^2}}$ as a multiple-valued function of the complex plane. We can construct its Riemann surface by cutting two copies of the complex plane along the line segment from $\displaystyle -1$ to $\displaystyle 1$ and pasting them together along this cut in the usual fashion. What we obtain is a cylinder, to which we could add two points at infinity (one point in each direction of the cylinder, i.e. one point on each of the two sheets); doing that, we'd obtain a surface isomorphic to the Riemann sphere, but let's stick to the non-compactified surface we had before (the cylinder).

    Now it's quite clear in view of the differential equation satisfied by $\displaystyle w=\sin z$ that $\displaystyle w$ has as inverse function the integral $\displaystyle z = \int_0^w {f(t)dt}$. However we all know $\displaystyle \arcsin w$ is far from a single valued function (unless we consider its image to be the cylinder $\displaystyle \mathbb{C}/2\pi \mathbb{Z}$, but let's not do that now). Therefore we can expect the multi-valuedness of the function $\displaystyle \arcsin w$ to be explained by the fact that the Riemann surface on which we are integrating has a non-trivial fundamental group; non homotopic paths of integration between $\displaystyle 0$ and $\displaystyle z$ should yield values of $\displaystyle w$ differing by an integral multiple of $\displaystyle 2\pi$. I've understood this to be true but I'm having trouble checking it computationally; I would like to check it.

    My question arises also in the context of elliptic function. I have a book in which Jacobi's $\displaystyle \mbox{sn}$ function is constructed as the inverse of the elliptic integral $\displaystyle \int_0^w \frac{dt}{\sqrt{(1-t^2)(1-k^2t^2)}}$; the author states that integrating around the branch points in various ways results in paving the plane with the period lattice of $\displaystyle \mbox{sn}$, but does not show it by a computation.
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    MHF Contributor Bruno J.'s Avatar
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    I believe I figured it out. We fix a sheet of $\displaystyle f(z)$ and we integrate $\displaystyle f(t)dt$ on that sheet. The sheet has a cut from $\displaystyle -1$ to $\displaystyle 1$ and therefore the integral will be defined up to an integral number of cycles around the cut. But the integral along a simple closed curve whose interior contains the cut is equal to the integral along the cut itself, by Cauchy's theorem (since whatever branch of $\displaystyle f(z)$ which we have is analytic in the region between the two curves). The integral along the cut is $\displaystyle \int_{-1}^{1}f(t)dt + \int_{1}^{-1}(-f(t)dt) = 2 \int_{-1}^1 f(t)dt = \pm 2\pi$, which is indeed a generator for the group of periods of $\displaystyle \sin z$.
    Last edited by Bruno J.; Apr 26th 2010 at 09:21 PM.
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