$\left.\frac12i\;\text{Log}\frac{1-(-i+e^{it})}{1+(-1+e^{it})}\right|_0^{2\pi}=\frac12i\left(\log\left|\frac{i}{i}\right|+i\arg 1-\log|1|-i\arg1+2\pi ik\right)$
could someone explain how this is equal, and how we get from the LHS to the RHS.
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Welcome to Math SE ! Do you know the logarithms of complex numbers ? – Claude Leibovici Nov 20 '14 at 15:10
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to some extent yes – smith Nov 20 '14 at 15:17
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$\log(z) = \log|z|+i\arg z$. – GEdgar Nov 20 '14 at 15:41
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would I input the 2 limits before applying the complex logarithm equation? – smith Nov 20 '14 at 15:51
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@Timbuc, hi I'm struggling with the same question as http://math.stackexchange.com/questions/1029158/circular-contour-integration , however, due to my low rep I cannot comment on the answer you've given, I was wondering if I could get some help with the question. I've fully read and understood your answer, however, have noticed that all answers will simply cancel down to $- \pi k$ .. I feel like I'm missing something? – smith Nov 20 '14 at 16:38
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@Timbuc , also could you look at http://math.stackexchange.com/questions/1030968/complex-logarithms , that is another part of the same question, part 1 – smith Nov 20 '14 at 16:45