Vanishing residue says the coefficient of the -1 term in the series expansion is 0( since it is 1/z^2), which doesn't imply the singularity is removable.
Hello everyone, I am having a little problem with the following integral:
Where D is the Domain D = {z in the complex field : |z|<2}
Now, my understanding is I have to look where my function goes to zero.
On the denominator I have :
z = j (simple pole)
z = -j (simple pole)
and z = 0 (pole of order 3)
Now, on the numerator, I have
z = 0,pi, 2pi
Since zero appears both at the numerator and the denominator I understand it becomes a pole of 2order, and not a 3 order one.
So my integral becomes:
Now what happens is, if I calculate the residue in ZERO it's ZERO! So it means that That's a removable singularity! Now, I double checked with wolfram and it doesn't even bother to calculate the residue in zero.
Could anyone tell me why? Please?
sinz/(z^3(z^2+1)) residue - Wolfram|Alpha
Thanks a lot!!
sin(z)/z would be a removable singularity. is, as you say, a pole of order 2. Yes, the residue at 0 is 0. That does NOT mean the the singularity is removable.
sin(z) is analytic and has McLaurin series
So the Laurent series for is
z= 0 is not a "removable singularity" because that series has negative powers. It is a pole of order 2 because the largest negative power is -2. The residue at z= 0 is the coefficient of the -1 power which is 0.