# Branch point problem and argument along contour

1. Mar 14, 2009

### jack5322

Hi everyone, I have a couple questions that have been gnawing at my head for some time:

Why is it that on wikipedia for the example logarithms and residue at infinity, (on the methods of conour integration link) that the integral is not zero?
there are no isolated singularities inside the contour, there fore by the residue theorem it should be zero.

Also, when i go around the countour of sqrt(z^2-1) i.e. the one encircling the two branch points in the counter clockwise sense, I always get that lim y to zero- is -isqrt(x^2-1) and y to zero+ is the negative of that. The answer though, is isqrt(1-x^2) and the negative of that for lim 0-. Why am i wrong?

Help would be greatly appreciated

Thanks!

2. Mar 14, 2009

### yyat

http://en.wikipedia.org/wiki/Method...80.93_logarithms_and_the_residue_at_infinity"
The function f is not holomorphic inside the contour, it has a discontinuity ("branch cut") along the interval [0,3], therefore the residue theorem does not apply.

I don't understand what you mean, could you elaborate on that?

Last edited by a moderator: Apr 24, 2017
3. Mar 14, 2009

### jack5322

well what happens is this:

we have the function sqrt(Z^2-1) we convert to polar coordinates i.e. sqrt(z+1)*sqrt(z-1)=r*p*e^i(theta_1+theta_2)/2 where theta's are the principle branch that is thetas are in beween zero and 2pi. With this choice of branch the line segment -1<x<1 is the cut line. why do we get the function equals what i was saying earlier on the above and below segments of the cut? that is, why does it equal i*sqrt(1-x^2) above the cut, and -i*sqrt(1-x^2) when we approach the cut from below

4. Mar 14, 2009

### jack5322

the contour used in evaluating the problem i'm looking at doesnt involve using the residue at infinity because there is a large circle around the cut line too. It therefore does not enclose the residue at infinity. why isnt there a large circle around the contour in the wikipedia example, it isnt consistent with the one in my book

5. Mar 14, 2009

### yyat

From what I know the principal branch of sqrt is cut at the negative real numbers, so theta is in the range from -pi to pi. This is consistent with saying that the cut of sqrt(Z^2-1) is along the interval [-1,1]. When writing sqrt(z^2-1)=sqrt(z+1)*sqrt(z-1) you need to be careful however, since you have to use different branches for the two sqrt on the right hand side to make the identity true, see http://en.wikipedia.org/wiki/Square_root#Notes".

Last edited by a moderator: Apr 24, 2017
6. Mar 14, 2009

### jack5322

i can understand that that is very important information, but for some reason my book uses the branch theta from 0 to 2pi but what would the function be if we took the limit from above and below and why? i'm very puzzeld by this

7. Mar 15, 2009

### jack5322

in one example they use the poles inside a larger contour with a the dogbone enclosed in a circle, this turns out to not use the residue at infinity but it still is correct while wikipedias does, why is this?