If you have a function of the form f(z)z^(1/2) and f(z) has a pole somewhere at z = p, then the residue at z = p clearly depends on how you define z^(1/2).
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?
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?
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.Actually that doesnt work. My question is this: Why is it that we calculate the residues from the negative square root if the 2pi is on top?
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.ok, why is it that the 2pi is on the top and the zero on the bottom for the example of something like (1-z)^-1/2 for -1<x<1 but the 2pi is on the bottom for the sqrtz with 0<x<infinity and zeroon the top?