Analytic continuation of ln(z) counterclockwise about the unit circle, 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,
 A: This problem exemplifies the necessity of taking a branch cut of $\mathbb{C}$ in order to define $\log z$. 
We begin with the case $\theta = 0$, which leads to the power series expansion $\log z = (z-1) - \frac{(z-1)^2}{2} + \dots$. Here, we consider $\log z$ under the principal branch cut $-\pi < \arg(z) < \pi$.
This expansion only converges inside of the disc $\left |z-1 \right | < 1$, and we wish to analytically continue our formulation of $\log z$ to a larger domain. 
To do this, we make 12 ``shifts", as you have written. First, consider $\theta = \pi/6$.  Using the formula 
$$a_n = \frac{1}{n!} \frac{\partial^n}{\partial z^n} \log (z) \bigg |_{z=e^{i\theta}},$$
it is easy to obtain a new power series expansion for $\log z$, which converges on a disc of radius $1$ centered at $z= e^{i\theta}$ by Hadamard's formula. 
Now, in order for this to be an analytic continuation of the previous power series (centered at $z= 0$), we must have agreement on the two discs of convergence. This can be seen by choosing the branch cut of $\log$ to be $-5\pi/6 < \arg(z) < 7 \pi/6$, which can be visualized as rotating the cut counter-clockwise by $\pi/6$ radians. What we now have is the analytic continuation of $\log z$ on $\{ z: 0 \leq \arg(z) < \pi/6, |z| < 1\}$. Actually, the full domain of the analytic continuation we have produced is
 $\{ z: \left |z-1 \right | < 1 \} \cup \{ z: \left |z-e^{i \theta} \right | < 1 \}$, but we choose this smaller domain so that we won't run into trouble in the last shift (explained in the next paragraph). 
Continue this process for $z= e^{in \theta}$, for $n = 1, 2, \dots, 11$, so that we have the analytic continuation of $\log z$ on $\{ z: 0\leq \arg(z) < 11\pi/6, |z| < 1\}$. Observe that while the power series expansion for $\log z$ centered at $z= e^{i 11\theta}$ converges on a disc of radius 1 centered at $e^{i 11\theta}$, this disc overlaps with the domain given by the power series expansion centered at $z= 1$, and so it would be impossible to choose a suitable branch cut that doesn't intersect the domain. 
Now, in each of these shifts, we have been able to expand the domain of $\log z$ so that $\log (e^{in \theta}) = in\theta$, for $n = 0, 1, \dots, 11$. Thus, if we are able to make the 12th shift, a power series expansion centered at $z = e^{i 12\theta} = e^{2\pi i} = 1$, we must have $\log(e^{i 12 \theta}) = 2\pi i = \log(1) + 2\pi i$. 
