Complex contour integral of square-root

  • Thread starter Yeggoua
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  • #1
Yeggoua
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For my homework I am told: "Evaluate $z^(1/2)dz around the indicated not necessarily circular closed contour C = C1+C2. (C1 is above the x axis, C2 below, both passing counter-clockwise and through the points (3,0) and (-3,0)). Use the branch r>0, -pi/2 < theta < 3*pi/2 for C1, and the branch r>0, pi/2 < theta < 5*pi/2 for C2."

I am unsure how to approach this and would appreciate help. I get that I need to take the anti-derivative to get 2/3*z^(3/2) and possibly convert to polar form, but am unsure how to handle the different 'branches' to find an answer.

Thanks
 

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  • #2
NateTG
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Yeggoua said:
For my homework I am told: "Evaluate $z^(1/2)dz around the indicated not necessarily circular closed contour C = C1+C2. (C1 is above the x axis, C2 below, both passing counter-clockwise and through the points (3,0) and (-3,0)). Use the branch r>0, -pi/2 < theta < 3*pi/2 for C1, and the branch r>0, pi/2 < theta < 5*pi/2 for C2."

I am unsure how to approach this and would appreciate help. I get that I need to take the anti-derivative to get 2/3*z^(3/2) and possibly convert to polar form, but am unsure how to handle the different 'branches' to find an answer.

The bit about the branches is that you need to choose which square root you use when you're integrating. (Recall that every number has two square roots.) That's really all that the 'branches' stuff is about.

You should be able to integrate along the two pieces seperately without any special problems.
 

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