A problem about residues(is matlab wrong?)

1. Dec 12, 2007

oahsen

I have written this simple code fragment to the matlab for finding the residue(s) of the function 1/(z-i/9)^3;

b=[1];
a=[1 -i/3 -1/27 i/(729) ];
[r p k]=residue(b,a)

and get the following result;
r =

0
0
1

p =

0.0000 + 0.1111i
0.0000 + 0.1111i
0.0000 + 0.1111i

k =

[]

The poles are true. Function has a pole of order three at z=i/9 . However, there are two different values for the residue: 0 and 1. I could not get the meaning behind that? How can a function have different residue at the same point? Additionally if we assume matlab is right then the result of the contour integral should be 2pi*i at the unit circle. However, we can easily show that this contour integral is zero. So, is something wrong with the residue algorithm of matlab or there is something I do not consider?

2. Dec 12, 2007

Avodyne

The residue of a pole in a function f(z) at z=a is the coefficient of 1/(z-a) in the Laurent expansion of f(z). In your case, a=i/9 and the coefficeint of 1/(z-i/9) is zero. So the residue (there is only one residue!) is zero.

3. Dec 13, 2007

oahsen

Yes, logically it should have only one residue which is zero. However, what I did not understand is why has matlab find also a residue which is 1 at the pol z=i/9 ?

4. Dec 13, 2007

Avodyne

I don't use matlab so I can't help.