- #1
Lyuokdea
- 154
- 0
I'm looking at an integral which in part involves finding the residue of [tex]\frac{1}{1+z^{n}}[/tex] at [tex]z=e^{\frac{i\pi}{n}}[/tex]
I thought the general method for residues was to find the 1/z term in the Laurent series, (which I'm not particularly sure how to do in this case), however, the answer provided does:
[tex]\frac{1}{\frac{d}{dz}(1+z^n)} = -\frac{1}{nz^{n-1}}[/tex]
evaluating this at [tex]z=e^{\frac{i\pi}{n}} [/tex], they obtain:
[tex]Res=-\frac{e^{\frac{i\pi}{n}}}{n}[/tex]
Why is this a legitimate method, for solving residues, can you always employ this method (it seems much easier), are there any other important methods for determining the residues of various things?
Thanks,
~Lyuokdea
I thought the general method for residues was to find the 1/z term in the Laurent series, (which I'm not particularly sure how to do in this case), however, the answer provided does:
[tex]\frac{1}{\frac{d}{dz}(1+z^n)} = -\frac{1}{nz^{n-1}}[/tex]
evaluating this at [tex]z=e^{\frac{i\pi}{n}} [/tex], they obtain:
[tex]Res=-\frac{e^{\frac{i\pi}{n}}}{n}[/tex]
Why is this a legitimate method, for solving residues, can you always employ this method (it seems much easier), are there any other important methods for determining the residues of various things?
Thanks,
~Lyuokdea