Probably obvious complex analysis question

  • Thread starter Mystic998
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  • #1
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Homework Statement



[tex]\int_{|z| = 2} \sqrt{z^2 - 1}[/tex]


Homework Equations



[tex]\sqrt{z^2 - 1} = e^{\frac{1}{2} log(z+1) + \frac{1}{2} log(z - 1)}[/tex]

The Attempt at a Solution



Honestly, my only thoughts are expanding this as some hideous Taylor series and integrating term by term. But I know there has to be a simpler way to go about it. I tried to find an antiderivative (with a little Calculus 2 cheating), but I got [itex]\frac{1}{2}(z\sqrt{z^2 - 1} - log(z + \sqrt{z^2 - 1}))[/itex], which may or may not be correct (I took the derivative, saw the horrendous expression that came out, and my brain immediately shut down). But even if it is, choosing an appropriate branch for the log function seems like it would be kind of tricky.

Anyway, suggestions would be appreciated. Thanks.
 

Answers and Replies

  • #2
NateTG
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Trig substitution comes to mind.
 
  • #3
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That's how I got the "antiderivative" I mentioned, but if that's right then I really have no idea how to choose the right branch of the logarithm. Do I have to see what [itex] z + \sqrt{z^2 - 1}[/itex] does to my path?
 
  • #4
olgranpappy
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Homework Statement



[tex]\int_{|z| = 2} \sqrt{z^2 - 1}[/tex]


Homework Equations



[tex]\sqrt{z^2 - 1} = e^{\frac{1}{2} log(z+1) + \frac{1}{2} log(z - 1)}[/tex]

The Attempt at a Solution



Honestly, my only thoughts are expanding this as some hideous Taylor series and integrating term by term. But I know there has to be a simpler way to go about it. I tried to find an antiderivative (with a little Calculus 2 cheating), but I got [itex]\frac{1}{2}(z\sqrt{z^2 - 1} - log(z + \sqrt{z^2 - 1}))[/itex], which may or may not be correct (I took the derivative, saw the horrendous expression that came out, and my brain immediately shut down). But even if it is, choosing an appropriate branch for the log function seems like it would be kind of tricky.

Anyway, suggestions would be appreciated. Thanks.

Let me see if I am interpreting your integral symbol correctly: It's a closed line integral around the circle |z|=2 in the complex plane. Right? Which would be zero if the integrand was analytic, but it's not (the value is actually something like [itex]\pi i[/itex], i think) so we should consider the types of singularities of the integrand... I believe I know how to do this integral so I'll be a little Socratic now, even though it pains me to be so:

The integrand has branch points, where are they?
 
  • #5
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You interpreted correctly.

Anyway, I think the branch points all lie on the real axis less than or equal to 1, assuming we're using the principal branch of the log.
 
  • #6
olgranpappy
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You interpreted correctly.

Anyway, I think the branch points all lie on the real axis less than or equal to 1, assuming we're using the principal branch of the log.

Well, you're right that the branch points of
[tex]
\sqrt{z^2-1}
[/tex]
all lie on the real axis... But where, exactly?
 
  • #7
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Well, log(z) has its branch points for z less than or equal to zero, so we get branch points for [itex]log(z^2 - 1)[/itex] whenever [itex]z^2 - 1 \leq 0[/itex]. So I guess it's [itex] -1 \leq z \leq 1[/itex]?
 
  • #8
olgranpappy
Homework Helper
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just look at the given function
[tex]
\sqrt{z^2-1}\;.
[/tex]

The polynomial under the square root can be factored very easily to see where the two branch points are.
 
  • #9
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I don't quite understand what a branch point is apparently. I mean, as I understand it, yes, -1 and 1 are branch points, but I thought that any point where the multi-valued "function" actually started to be multi-valued was a branch point.

Oh well, -1 and 1 is the answer you're looking for I suppose.
 
  • #10
olgranpappy
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yes, the function of interest has two branch points: one at -1 and one at +1. It also has a simple pole at infinity. An appropriate branch cut may run along the real axis from -1 to +1 (connecting the branch points). The contour (|z|=2) can be deformed without changing the value of the integral by shrinking it down without crossing the cut. we can shrink it down very far until it just wraps around the cut. the value of the integral is twice the integral just above the cut from -1 to +1. I.e., [itex]2 \int_{-1}^{1}dx \sqrt{x^2-1}=i\pi[/itex].
 
  • #11
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Well, we haven't officially discussed path homotopy and its implications for contour integration yet (much to my chagrin), but I think I'm getting the idea. I think I could look at [itex]log(\frac{z+1}{z-1})[/itex], then maybe use that to define a square root in the region I want.

Anyway thanks.
 

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