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Using Morera's Theorm

  1. Sep 29, 2012 #1
    The problem states: Use Morera's Theorem to verify that e^z is indeed analytic inside a circle of radius R.


    Morera's Theorem states: If f(z) is continuous in a domain D and if ∫f(z)dz=0 inside any closed curve in D, then f(z) is analytic in z.


    I showed that e^z is continuous inside a circle of radius R. Then I showed that ∫f(z)dz=0 over any circle of radius R inside D. I don't know how to show that this applies to ANY closed curve in D. I'm trying to do something with cross cuts. I let C be a circle for which I know the integral is zero. I then introduce cross cuts: I let Co be the curve for which I know the integral is 0. Then I cut it somewhere (call the cut L1) and L1 connects to another curve γ (the curve I'm trying to deform C to). Then there is another curve from γ back to Co. In the limit L1=-L2 so they cancel. And I'm left with Co-γ (gamma goes in the opposite direction of Co). The integral over Co is also 0 since in the limit it is the same as C.

    The problem I'm having is showing that the integral over γ is 0.
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Sep 29, 2012 #2
    Hmmm That's hard! Sorry, I forgot my complex analysis, and I'm EE so I'm not exactly the best at math, but this problem is interesting so I'll give it a thought.

    So that gamma curve you mentioned, which you are having trouble showing is zero, I think encloses everything outside the curve, instead enclosing an area inside the curve (since it is in the opposite direction). That means it is essentially the same as an infinite circle enclosing e^(1/z). So maybe you can show that for all R≠0, ∫e^(1/z)dz =0 (that should be the same process as for ∫e^(z)dz). Then you can just take lim R→∞ ∫e^(1/z)dz =0, which means your gamma curve is zero???

    Sorry I can't be more help, best of luck!
     
  4. Sep 29, 2012 #3

    Dick

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    I'm guessing you only proved that the integral around circles centered at the origin is zero. I don't think your program is going to work. If you pick f(z)=|z|e^z then the integral around circles centered at the origin is also zero. But f(z) isn't analytic.
     
  5. Sep 29, 2012 #4
    I thought taking a circle around zero would be good enough, because if I showed that ∫(e^z)dz=0 over a circle, and then showed by way of deformations that it was also true for an arbitrary curve γ, then I could conclude it was true for ALL curves inside of C. So taking C large enough to bound the domain, and then deforming it to an arbitrary curve in the domain.

    I was expecting gamma to be inside of C. I figured that gamma was negative because I was taking cross cuts and it reverses on the inside.

    Perhaps a picture would explain what I mean better.

    *I have tried to attach a picture, but I'm very new and I'm not sure if it worked*

    In the picture the graphic on the left shows a Curve C being deformed to a curve Co with cross cuts and smaller circles around the singularities (the x's). Then the right picture shows in the limit the cross cuts cancel each other. It goes on to use the analyticity of f(z) to say that the integrals over deformed curves are equal to eachother.

    This is the part I'm struggling with: all the theorems in my book use Cauchy's Theorem (∫f(z)dz=0 around a closed curve C with f analytic on the interior of C) or other theorems assuming the analyticity of f(z). But I can't use these since this is what I'm trying to prove.
     

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  6. Sep 30, 2012 #5
    Yeah as you were saying, all those theorems involve analytic functions. That picture is actually using deformation theorem (don't know formal name) to show that any closed contour around the singularities is the sum of the residues. However, you cannot use this since deformation theorem requires a function to be analytic. So I think, all the proofs you see using cuts, already require the function be analytic, which has certain properties that allow you to deform the curve (i think from Cauchy Goursat). So in your case you want to deform curves to show the function is analytic, but deformation requires function be analytic. So in short I think you might be headed in the wrong direction.

    But anyway, as I said earlier, I am not nearly qualified, maybe try emailing professor, he can give you a hint. Either way, it seems really hard so good luck with it!
     
  7. Sep 30, 2012 #6
    I think I got it....hint: product of two analytic functions is analytic!
     
  8. Sep 30, 2012 #7
    I don't see it yet! I just don't understand how I can show the integral of a function is zero over ANY curve (without using Cauchy's Theorem, or explicitly using the anti-derivative).

    Although I guess continuity guarantees that there's an anti-derivative?
     
  9. Sep 30, 2012 #8
    No continuity does not imply antiderivative (Im pretty sure at least!), otherwise you'd have analytic (since then the antiderivative would be analytic and therefore the derivative of the antiderivative (the original function) must also be analytic...analytic is a very strong condition.

    As I mentioned earlier, since you trying to prove this thing the other way around, trying to do this the way you mentioned earlier will get you in an infinite conundrum! Its like proving a theorem with that same theorem...not possible!

    So anyway the hint I gave you will allow you to prove it for ANY curve, just think about it, your professor (or whoever assigned the problem) gave you e^z for the reason. So I said the product of two analytic functions is analytic, Can you think of e^z as a product of two functions?
     
  10. Sep 30, 2012 #9
    it actually has 5 parts. e^z, z^n for n≥0, sinz, lnz (on the principal branch), then what can I say about sinz/z using Morera's theorem.

    I can't think of any complex functions. Maybe [itex]e^{x}[/itex](cosy+isiny)

    cos(z) and sin(z) don't work. cosh(z) and sinh(z) is closer but I still don't see it. Maybe the sum of analytic functions? Then cosh(z)+sinh(z) works.

    I still don't see how this helps me show the integral over any curve is 0 (unless I use Cauchy's Theorem on the other analytic functions that compose e^z?)

    But that doesn't seam very satisfying to me.
     
  11. Sep 30, 2012 #10
    Yes you're last idea is the right idea. In a sense, Cauchy and Morera are essentially saying the same thing, especially in this case since we know e^z has no singularities.
    You're thinking too complex about the function decomposition... z= x + iy, so e^z = e^x*e^iy.... those two functions can very easily verified as analytic by Morera's , then use product of two analytic functions is analytic...and you're done!
     
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