We write ln(z) as ln(1+z-1) = ln(1+(z-1)) to utilize the familiar expansion that is:
(z-1) - (z-1)^2 / 2 + ...
which converges for |z-1| < 1, i.e., we get convergence of ln(z) in an open Taylor disk centered at z = 1, with the boundary being a circle of radius 1.
Now I want to analytically continue ln(z), along $z=e^{i\theta}$, that is, continue ln(z) along the unit circle |z| = 1, with expansions / Taylor disks centered at $e^{i\theta}$.
I am able to write out explicitly the new power series in powers of (z-$e^{i\theta}$), with new coefficients, and noting that we now have convergence for |z-$e^{i\theta}$| < r < 1.
My question is: Say, on my first shift, shifting the center from z = 1 to z = $e^{i(\pi/6)}$, I get a new power series, but here's my confusion: is this an expansion for the original ln(z), with the original, chosen branch cut, say, the negative real axis? I'm guessing it can't be, because as I continue analytically around the unit circle, I will eventually get a Taylor disk that covers a part of the negative real axis, which would be nonsense. If indeed, each shift requires me to specify a new branch cut, how do I do this, so that we still have convergent Taylor disks that satisfy |z-$e^{i\theta}$| < 1?
(My task is to do 12 shifts, starting at $\theta = pi/6$ to get back to z = 1 and that I should be able to observe that I get back the original ln(z) plus an additional $2\pi i$.)
Edit: But if I chose a different branch cut, as I shift the center and re-expand the power series, then this new series wouldn't even be an analytic continuation of the original ln(z), so I don't know how to proceed.
Thanks,