Holomorphic 1/z

 Spoiler

Interestingly, WolframAlpha says that the integral is θ+i*log(1+3exp(iθ). Then we run into the problem of taking the log the form that I was having trouble with earlier, namely the one I bolded and inferred:

log(3exp(4iπ+1) - log(3exp(0i)+1) = 4πi

but e^(4 pi i) = e^(2 pi i) = e^(0 pi i) = 1, so log(3e^(4 i pi) + 1) = log(3 + 1) = log(3 e^(0 pi i) + 1), and that leaves us with zero?

This is precisely what I did, without shifting the "index" (aka I parametrized every line starting from (1,1) from 1 to -1 and then from -1 to 1 when I reached (-1,-1). I do see the problem now, however. In the last step, simplification had to be grouped:

∮1/z dz = ∫ 1/(t+i) dt [1,-1] + ∫ i/(-1+it) dt [1,-1] + ∫ 1/(t-i) dt [-1,1] + ∫ i/(1+it) dt [-1,1]
= ln(t+i) [1,-1] + ln(-1+it) [1,-1] + ln(t-i) [-1,1] + ln(1+it) [-1,1]
= {ln(1+i) - ln(-1+i)} + {ln(-1+i) - ln(-1-i)}+ {ln(-1-i) - ln(1-i)} + {ln(1-i) - ln(1+i)}
= ln[(1+1)/(-1+i)] + ln[(-1+i)/(-1-i)] + ln[(-1-i)/(1-i)] + ln[(1-i)/(1+i)]

And you claim that each one of these grouped logarithms is equivalent to πi/2.

No, I claim that specific result to be -2 pi i as you went clockwise instead of counterclockwise.

Also, Wolfram Alpha is partly correct, but look up complex logarithms! Some of them have infinitely many solutions.

Noah