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

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The integral of 1/(z-2) - 1/(z-1/2) is evaluated using the residue theorem, which relates to the integral of 1/z around a closed contour. The integral of 1/z over a circle is known to equal 2*pi*i, and the discussion focuses on how this applies to the given expression. The result of the integral is calculated as 1/3 times the sum of residues at the poles, leading to -2*pi*i/3. The connection between the integrals is clarified through the properties of complex functions and their singularities. Understanding these relationships is crucial for solving similar integrals in complex analysis.
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))??
 
Question: A clock's minute hand has length 4 and its hour hand has length 3. What is the distance between the tips at the moment when it is increasing most rapidly?(Putnam Exam Question) Answer: Making assumption that both the hands moves at constant angular velocities, the answer is ## \sqrt{7} .## But don't you think this assumption is somewhat doubtful and wrong?

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