A problem about residues(is matlab wrong?)

[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?

 

[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.

 

[oahsen]

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.

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 ?

 

[Avodyne]

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