Holomorphic 1/z

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

However, I'm half-willing to believe that the integral is 0, since it doesn't have any singularities, so it should be holomorphic. But then again, that would mean that 1/(1+3exp(-iθ) is also holomorphic, and you could have simplified it the other way and obtained the wrong solution.

Now, the contour integral will look like this:



The first integral evaluated is this:



The remaining integrals return the same value (which should not be a surprise, as this figure is symmetric!), and is up to the reader to evaluate, as I am too lazy to do it. The value of the contour integral is thus 2 pi i.

This pretty much 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 must be grouped before evaluated:

∮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+i)/(-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. Turns out to be -2πi. You're right. I seem to have gone backwards. I'll check my parametrization later.

You guys are giving me too much credit, I don't know everything!

You seem to be beyond us all on math. I've yet to observe a math post that you couldn't answer, on Southperry, at least.