Is there a pole at z=-1 in this complex integral problem?

[Topher925]

I've come across an complex integral that I just can't seem to figure out. Basically I need to integrate f(z) = 1/(z^6-1) around the circle |z+1|=1.

At first glance the radius of the circle must be zero in order to satisfy |z+1|=1 and therefore the function, f(z), is analytic in D so therefore the integral is zero. However, I've got the feeling this isn't correct. Am I doing something wrong?

 

[Citan Uzuki]

The circle is of radius 1, centered at z=-1. Incidentally, this means that you have two poles of the function on the circle, which will definitely cause problems in integrating. Maybe you could take some kind of Cauchy principal value?

 

[hunt_mat]

Th way to deal with pole on the contour is to do an indentation around the points then when you come to do the integral you have to calculate the integral on the indentations.

 

[jackmell]

I've come across an complex integral that I just can't seem to figure out. Basically I need to integrate f(z) = 1/(z^6-1) around the circle |z+1|=1.

At first glance the radius of the circle must be zero in order to satisfy |z+1|=1 and therefore the function, f(z), is analytic in D so therefore the integral is zero. However, I've got the feeling this isn't correct. Am I doing something wrong?

If I may try to also be of some assistance. First need to get a few things straight like that absolute value thing. If z=-1+e^{it}, unit circle around the point -1 right, then z+1=e^{it}, take the absolute values then |z+1|=1. Also, in Complex Analysis, you need to always be alert for principal-valued integrals. They're common and need to approach problems expecting to encounter them. First though, look at a simple one:

\int_{|z|=1} \frac{dz}{z-1}

Get that one straight first before you do yours. That expression is an implicit principal-valued integral. We could have written it as:

\int_{|z|=1} \frac{dz}{z-1}\rightarrow \text{P.V.} \int_{|z|=1} \frac{dz}{z-1}=\int_P

Now form the circular contour around the origin with an (exterior) indentation around the point z=1 so that by the Residue Theorem:

\int_P+\int_I=2\pi i r

So principal-value+indentation equals 2 pi i times sum of residues right? So that we have:

\int_P=2\pi i r-\int_I

Then look up that theorem about integrating over an indentation around a simple pole as it's radius goes to zero so that we can write:

\int_P=2\pi i-\pi i=\pi i

and here's how to check it in Mathematica:

In[5]:=
NIntegrate[(1/(z - 1))*I*Exp[I*t] /. z -> Exp[I*t], 
  {t, -Pi, 0, Pi}, Method -> "PrincipalValue"]

Out[5]=
0. + 3.1415926535897647*I

Now do yours and then see if you can code the check in Mathematica.

 

[Topher925]

Thanks for all the help. I think my biggest problem was just not realizing that there was a pole at z = -1. Guess I should stop doing math when I'm really tired. Anyway, I was able to find the solution using Residue Integration as stated above. Once again, thanks for the help, I'm quite terrible at math.