# Complex integral

1. May 20, 2006

### Zurtex

Hey,

I'm about to do an exam tomorrow and I seem to be a little stuck on how to answer this problem from a past paper.

We've mainly been integrating by using the residue theorem, I forgot how to do something like this:

$$\int_{C} \left( z^2 - 1 \right)^\frac{1}{2} dz$$

$$C = \{ z : |z - 10| = 25\}$$

Any help would be much appreciated.

2. May 20, 2006

### TD

I'd say parametrize C, which is a circle with center (10,0) and radius 25, let x = cos(t)+10 and y = sin(t) and use this in z = x+iy. Not sure though

3. May 20, 2006

### vsage

It's been a year since I've had to work a Residue Thm problem, but I don't see any poles at all in that problem. IIRC that means there is no residual term?

Last edited by a moderator: May 20, 2006
4. May 20, 2006

### TD

Actually, I think you're right. Our 'approaches' should give the same result, but let's stick to yours unless you want to calculate an annoying integral

From the Cauchy-Goursat theorem, I believe this should be 0

5. May 20, 2006

### ircdan

Yea it's 0 because the integrand is analytic inside and on that circle, so it follows from what you said. :)

Last edited: May 20, 2006
6. May 20, 2006

### mathwonk

i am a little puzzled by this claim as the integrand is not singlevalued inside the path.

i.e. the square root is singlevalued only on paths that contain both 1 and -1 in their interior.

7. May 21, 2006

### Tide

One way to do the integral is to recognize that it is equivalent to the integral about a contour enclosing both branch points (z = 1 and z = -1). For convenience, choose to set the branch line as a straight line connecting the two branch points.

Thus, your new contour consists of two straight lines (one just above Im z = 0 and one just below) along with two infinitessimal loops about the branch points. You should be able to convince yourself that the infinitessimal loops contribute nothing to the integral (as noted above they are not poles).

Along the upper straight line segment, set $z - 1 = \lambda e^{i \pi}$ and $z + 1 = (2 - \lambda) e^{i 0}$ and, along the lower portion, set $z + 1 = \lambda e^{i \pi}$ while $z - 1 = (2 - \lambda) e^{i \pi}$. The rest is bookkeeping making sure you set the limits of integration correctly along with the signs of $d \lambda$

Last edited: May 21, 2006
8. May 21, 2006

### Zurtex

Hey, I didn't really understand any of that,

A friend of mine showed me the solution that was given by the lecturer:

$$\left( z^2 - 1 \right) = z \left( 1 - \frac{1}{z} \right)^\frac{1}{2}$$

Then expanded that in terms of the laurent series and got the residue about 0 as being -1/2 so the answer:

$$\int_{C} \left( z^2 - 1 \right)^\frac{1}{2} dz = -\pi i$$

Written below in her notes it has something which the lecturer said:

Q: Why singularity point z =\= +-1, but z=0?
it's not isolated singularity at point +-1 the equation is 0^(n^1/2) which is not analytical
therefore not included to work on

Oh well, doesn't make much sense to me, but all I need to do is pass an exam, not make sense.

9. May 21, 2006

### Hurkyl

Staff Emeritus
Here's another idea. Your function is analytic outside of your contour, right? What is its residue at infinity?

(Or, do a change of variable z --> 1/w, so you don't have to think in projective space)

10. May 21, 2006

### mathwonk

thats the way i looked at it. but remember to change dz to d(1/w) which is not dw.

11. May 22, 2006

### Zurtex

Thanks a lot people.

This when I am starting to wish they taught Complex Analysis rather than Complex variables. We don't get taught any of this from first princaples, just all the rules and how to do them, very annoying for when you want to know what is going on.

Oh well, nevermind, just had the exam and it went really well .

12. May 22, 2006

### mathwonk

in your example the region inside the circle of integration had two points where z^2-1 was equal to zero, hence two places where the square root function changes sign when a loop encircles one of those points. this emans that the square root function is really defined on a "double cover" of that circular region with "branch points" at both 1 and -1.

a double cover of a disc with two branch pionts is not so easy to picture, but is not a disc. It is a cylinder, and your integral was taken around only one end circle of the cylinder. cauchy's thm says if the integral is taken over the full bondary of a surface, it is zero, but all it says here is the integral is minus the integral over the other boundary circle.

i.e. if w = sqrt(z^2-1), then w^2 = z^2-1 defines a curve in the plane on which the function w really is well defined. but that curve projects via the projection taking (z,w) to z, two to one onto the z plane. since there is only one point over each of the points z = 1 and z = -1 we call those branch points.

the cylindrical region mentioned above, is the inverse image under this projection, on the curve with equation w^2 = z^2 -1, of the interior of your circle of integration in the z plane. abel and galois studied these integrals, but since the both died in their 20's, riemann was first to lay it all out. he of course lived to the ripe old age of 39 or so.

this geometric analysis does not make it any easier to calculate your integral, but does show what is going on.

people think riemann surfaces are very fancy, but all they are is essentially the graph of the (multi valued) function. i.e. the graph of a single valued fucntion apsses the vertical line test, and the graph of a general multivalued one does not, thats all.

i.e. the "riemann surface" of the multivalued function w = sqrt(z^2-1) is just the graph of the equation w^2 = z^2 -1.

there is a little more to it, like adding in points at infinity, if any, and then removing singular points, if any, but those are technicalities.

they do affect the genus though as hurkyl discovered for an elliptic curve.

i.e. the riemann surface of the function sqrt(z^3-1) is the graph of w^2 = z^3-1. completed by adding one point at infinity, and has genus one, given by the formula g = (1/2)(d-1)(d-2) where d is the degree.

for the function sqrt(z^4-1) however, the riemann surface is the graph of w^2 = z^4-1, but when you add in the point at infinity, it is not a manifold point, but a singular point and in this case analyzing it shows it subtracts 2 from the genus. so the genus is (1/2)(d-1)(d-2) - 2, where the dewgree is 4, so we get 3-2 = 1 again for the genus.

in your case, a double cover of a sphere (the z plane with one point at infinity) with 2 branch points is again a sphere, so the genus of the riemann surface of w = sqrt(z^2-1), i.e. the completed graph of w^2=z^2-1, has genus again (1/2)(d-1)d-2) = 0, where d = 2.

Last edited: May 22, 2006
13. May 22, 2006

### mathwonk

note that in your case we can consider the interior of your circle of integration separately from the exterior. both are disc in the riemann sphere. the inverse image of the interior disc is a cylinder ebcause there are two branch points in there.

the inverse image of the exterior disc is just 2 copies of a disc because there are no branch points out there. so the riemann surface is obtained by gluing one disc on each end of a cylinder, giving another sphere.
in hurkyl's examples, in each case there are 4 branch points, so if we write the riemann sphere as a union of 2 discs, the inverse image of a disc containing 2 branch points is a cylinder.

since there are 4 branch points, we can decompose the sphere into two discs each containing two branch points, so the riemann surface is a union of two cylinders glued at their end circles, i.e. a torus of genus one.

14. May 22, 2006

### Zurtex

I don't know what a "branch point" means, we were never taught that.

15. May 24, 2006

### mathwonk

it is a point where there are fewer than the expected number of values of your multivalued function, like 1 and -1 for sqrt(z^2-1). without them there would be no problem.

16. May 24, 2006

### mathwonk

there are two square roots of most numbers, i.e. except zero. so try to imagine the graph of the square root of z^2-1. it has two points iopvwer everyu point except 1 and -1.

here is how to build it: take two copies of the disc of radius 2, centered at 0.

cut both copies along the segmnent from -1 to 1. this gives you two cylinders, i.e. two slit discs.

then identify the top edge of the slit on one slit disc with the bottom edge of the slit on the other slit disc. this makes the two cylinders into one cylinder, but in such a way that it lies doubly over the original disc.

this gives a 2:1 cover of the original disc except there is only one point over each of the two points 1 and -1. thus thiose are "branch points".

i.e. there is a 2:1 map branched at 2 points from a cylinder to a disc.

thus the part of the graph of sqrt(z^2-1) lying over any disc containing the two branch points 1 and -1, is a cylinder.

it takes work to see this, but it is possible. you are younger and smarter than i am and i eventually became able to see it.

17. May 25, 2006

### Zurtex

I have no idea what that had to do with the integral though.

Here's a question I asked my lecturer when we taught methods on how to integrate in complex space (or contour integration or whatever it is called):

"What does it actually mean to integrate in complex space? Like real space is a bit like working out the area under the curve"

This is the answer I got:

"It doesn't mean anything"

All I know is rules and methods on how to integrate, words like branches and concepts like trying to describe what it actually looks like don't relate at all to how I've learnt to integrate.