MHB How to find the integral of a complex number

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The discussion focuses on calculating the integral of a complex function along different paths. The user is specifically stuck on the third segment of the integral from 1+i to i, where they compare their solution with another's. The calculations for each segment show how to evaluate the real part of z squared along specified paths, leading to a final result of 1 + i for the total integral. The user realizes that the difference in their solution was due to the orientation of the path. Understanding the orientation resolves the confusion regarding the integral's evaluation.
aruwin
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Hello.
I am stuck at the third point, that is from 1+i to i. I asked someone to show me his answer but that part of his is different from mine. Is his solution correct?
Here it is:

(i) z = 0 to 1 via z(t) = t with t in [0, 1]:
∫c1 Re(z^2) dz
= ∫(t = 0 to 1) Re(t^2) * 1 dt
= ∫(t = 0 to 1) t^2 dt
= 1/3.

(ii) z = 1 to 1+i via z(t) = 1+it with t in [0, 1]:
∫c2 Re(z^2) dz
= ∫(t = 0 to 1) Re((1 + it)^2) * (i dt)
= ∫(t = 0 to 1) (1 - t^2) * i dt
= i(t - t^3/3) {for t = 0 to 1}
= 2i/3.

(iii) z = 1+i to i via z(t) = t+i with t in [0, 1] and opposite orientation:
∫c3 Re(z^2) dz
= -∫(t = 0 to 1) Re((t+i)^2) * 1 dt
= -∫(t = 0 to 1) (t^2 - 1) dt
= -(t^3/3 - t) {for t = 0 to 1}
= 2/3.

(iv) z = i to 0 via z(t) = it with t in [0, 1] and opposite orientation:
∫c4 Re(z^2) dz
= -∫(t = 0 to 1) Re((it)^2) * i dt
= -∫(t = 0 to 1) -it^2 dt
= i/3.

So, ∫c Re(z^2) dz = 1/3 + 2i/3 + 2/3 + i/3 = 1 + i.
 

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Re: how to find the integral of a complex number

Aha, now I get it! It has an opposite orientation. I got it!
 

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