Why Poles move to LOWER Half Plane

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SUMMARY

The discussion focuses on the movement of poles in the complex plane when solving the Greens Function for the driven harmonic oscillator using the Residue method of integration. By adding a complex quantity (iy) to the angular frequency (w), the poles shift to the LOWER half plane, contrary to the initial assumption that they would move to the UPPER half plane. The integration of functions with cos(x) - E in the denominator illustrates the assignment of pole locations, specifically at x = cos^-1(E), which are positioned in the upper left and lower right portions of the complex plane.

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Master J
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Pretty simple question here guys, but I can't get my head around it.

When solving say, the Greens Function for the driven harmonic oscillator, one needs to use the Residue method of integration. So, one adds a complex quantity to your variable w (angular frequency), so we now have w ---> w + iy .

This moves the poles into the LOWER half plane. But why is that. I would have intuitately thought them to move to the UPPER half plane since we are adding and not subtracting a complex quantity.

Can someone shed some light on this?
 
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Perhaps I should restate my question...

Say I am integrating a function with cos(x) - E in the denominator, where E is a constant. So I have poles at:

cos(x) - E = 0

or

x = cos^-1(E)

so how do I assign the locations of these poles in the complex plane. I know that they are in the upper left and lower right portions, but I cannot see how this?
 
I can't tell if you are asking a completely simple question, but I only have an answer for a simple question, so here goes:

Say you have a graph of the function f(x) of a single real variable and you want to move that graph 1 unit to the right along the x-axis. Which function would you plot, f(x+1) or f(x-1)?

It's rather like daylight savings time adjustments. Spring forward, fall back. If want to move the graph a(x) of your daytime activity to the left in Spring so that you get up earlier then set your clock forward and use a(x+1).
 

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