- #26

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- Thread starter jack5322
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- #26

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- #27

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so the specified value of the multifunction is the same value on the top of the cut?

- #28

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Make some arbitrary choice for that and then the values are fixed everywhere. Why not try out a simple example, like the integral of sqrt(x)/[x^2 + 1] from x = 0 to infinity? In this case you can choose a contour from epsilon to R, half a circle to minus R and then from minus R to minus epsilon and then half a circle back to epsilon.so the specified value of the multifunction is the same value on the top of the cut?

- #29

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how do we know which part of the multifunction to compute for the residues?

- #30

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The choice is fixed when you define the function. If you define sqrt(z) and then want to evaluate the residue at z = i, there is no ambiguity, because you know what to choose for the polar angle for z = i.how do we know which part of the multifunction to compute for the residues?

- #31

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- #32

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why is it that we choose the negative sqrt if the 2pi is on top?

- #33

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You don't have to. Do it differently and see what happens.why is it that we choose the negative sqrt if the 2pi is on top?

- #34

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- #35

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- #36

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- #37

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In that case the polar angle goes from 2pi to 4 pi. When comnputing the residue, you have to evaluate a square root of some complex number. You write it in polar form, but then the angle will be between 2 pi and 4 pi.

- #38

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- #39

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If you have 1-z, then the line theta = 0 moves from z = 1 in te direction of z = 0. This is because you have to use the polar representation of 1-z. This is the only relevant difference between the two cases.

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