i think.... a bit confused because there will be singularities in UHP and LHP...

let z^4=z^2

now from here... how the hell do i do the residues.. the singularities are all over the place???

please someone help

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- May 27th 2010, 11:30 PMfredrick08another complex question

i think.... a bit confused because there will be singularities in UHP and LHP...

let z^4=z^2

now from here... how the hell do i do the residues.. the singularities are all over the place???

please someone help - May 27th 2010, 11:36 PMDrexel28
- May 27th 2010, 11:54 PMchisigma
The poles of the function are at , . Only the residues of the poles with positive imaginary part give contribution of the integral that is...

(1)

... where...

(2)

Kind regards

- May 28th 2010, 03:41 PMfredrick08
- May 29th 2010, 01:54 AMHallsofIvy
No, the limits are not all 0. If the numerator is 0 but so the denominator, is also 0. Use L'Hopital's rule to find the limits.

- May 30th 2010, 07:27 PMigopogo
I think there's a simpler way. Consider

where

and

Then , and for each of your singularities.

Then the residue at is

There are usually singularities in the UHP and LHP with these problems, but the LHP singularities are never inside the contour that you construct.