Why Does the Integral of 1/(z-2) - 1/(z-1/2) Equal -2*pi*i/3?

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SUMMARY

The integral of the function 1/(z-2) - 1/(z-1/2) evaluates to -2*pi*i/3 when integrated around a closed contour. This result stems from the application of the residue theorem, specifically utilizing the known integral ∫dz/z = 2*pi*i across a circle. The integral's value is derived from the residues at the poles z=2 and z=1/2, leading to the conclusion that the integral evaluates to 1/3 times the sum of the residues, resulting in -2*pi*i/3.

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Dassinia
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Hello,
I don't get why using the fact that ∫dz/z = 2*pi*i accros the circle
This integral gives:

1/3∫(1/(z-2)-1/(z-1/2))dz = 1/3(-2*pi*i) across the circle

Thanks !
 
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Dassinia said:
Hello,
I don't get why using the fact that ∫dz/z = 2*pi*i accros the circle
This integral gives:

1/3∫(1/(z-2)-1/(z-1/2))dz = 1/3(-2*pi*i) across the circle

Thanks !

Why do you think 1/z has anything to do with (1/3)*(1/(z-2)-1/(z-1/2))??
 

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