Contour deformation

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'Fraid my complex analysis is non-existant.

I would think a bit more carefully about the branch of the logarithm you're using there.

This is what I can't seem to get my head around for some reason.

I defined $\theta \in [0,2\pi)$ so the principal arguments are in this region, hence I'm not sure how I can get a $2\pi$ or any multiple of it, when it comes to evaluating the log of the complex number, as they would just go to the principal argument of 0.

Unless I say something like $\log(2e^{2\pi i}+1) = \log 3$ and $\log (2e^{0\pi i}+1) = \log 3 + 2\pi i$ but TBH I'm confused enough to not see where this would come from or why what I've written is not valid.

First thing to note is that there is a sign error here in the former equality, the relevance of which will become obvious in a moment. This second is that, by virtue of using the complex logarithm, you've made a choice of branch and consequently cannot expect both terms to evaluate to $\log 3$ as the two $\theta$s differ by a multiple of $2\pi$.


You're right in saying that you need something like the above, and the barrier appears to be arising from a misunderstanding of the principle value of the logarithm. Recall that the p.v. of the logarithm (with your convention of [0,2π) defining "principle" ) is:

$\text{Log } z = \log |z| + i\text{Arg }z$, where $0\leq \text{Arg } z < 2\pi$.

This is defined by a branch cut along the positive real axis, and from this definition it should be obvious that:

$\mathrm{Log}[2e^{i\theta}+1]_0^{2\pi} = (\log 3 +2\pi i) - (\log 3 + 0\pi i) = 2\pi i \not= 0$

After you correct the sign error above, this is then consistent with the result you obtained using the other circular contour.

One slightly subtle point about the arguments - as a consequence of the branch we've chosen to work with, if $z_+$ is 'just above' the positive real axis and $z_-$ 'just below' the positive real axis, we have $\text{Arg }(z_+) = 0, \text{Arg }(z_-) = 2\pi$. If you're confused by the fact that $\text{Arg }(z_-)$ is not in the interval of principle arguments, think about what happens to the argument as $\theta$ approaches $2\pi$ in the limit - in order to return to an argument of zero, we would have to cross over a discontinuity of the log and this can't happen.

Notice that, when $\theta=\theta_+$ is 'just above' $\ 0$, $z_+=2e^{i\theta_+} +1$ is 'close' to $z=3$ and 'just above' the real axis i.e. $\arg{z_+} = 0$. On the other hand, when $\theta=\theta_-$ is 'just below' $2\pi$, $z_- = 2e^{i\theta_-} + 1$ is still close to $z=3$ but 'just below' it i.e. $\arg{z_-} = 2\pi$.