Probably obvious complex analysis question

In summary, you attempted to find an antiderivative for the function z^2 - 1, but you got an expression that may or may not be correct. You also tried to find a branch of the logarithm that would be appropriate for the given function, but you were not able to find an answer.
  • #1
Mystic998
206
0

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.
 
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  • #2
Trig substitution comes to mind.
 
  • #3
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
Mystic998 said:

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
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
Mystic998 said:
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
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
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
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
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
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.
 

Related to Probably obvious complex analysis question

What is complex analysis?

Complex analysis is a branch of mathematics that deals with the study of complex numbers and their functions. It involves the use of complex variables and their properties to understand and solve problems in many areas of mathematics such as calculus, geometry, and physics.

What are complex numbers?

Complex numbers are numbers that have both a real part and an imaginary part. They are written in the form a + bi, where a and b are real numbers and i is the imaginary unit (√-1). Complex numbers can be plotted on a complex plane, with the real part being the x-coordinate and the imaginary part being the y-coordinate.

What is the difference between a complex function and a real function?

A complex function is a function that takes complex numbers as input and outputs complex numbers. On the other hand, a real function takes real numbers as input and outputs real numbers. Complex functions have more complex behavior and properties compared to real functions.

What is the Cauchy-Riemann equation?

The Cauchy-Riemann equation is a fundamental equation in complex analysis that describes the relationship between the real and imaginary parts of a complex function. It states that if a complex function is differentiable at a point, then its partial derivatives with respect to the real and imaginary parts must satisfy a set of equations known as the Cauchy-Riemann equations.

What are some applications of complex analysis?

Complex analysis has many applications in mathematics, physics, and engineering. It is used in the study of fluid dynamics, electromagnetism, signal processing, and many other fields. It also has practical applications in areas such as image processing, control theory, and quantum mechanics.

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