Analytic continuation of the logarithm This is an example from Serge Lang Complex Analysis book, it says


  Let us start with the function $log(z)$ defined by the usual power
  series on the disc $D_0$ which is centered at $1$ and has radius $&lt;1$
  but $&gt;0$. Let the path be the circle of radius $1$ oriented
  counterclockwise as usual. If we continue $log(z)$ along this path,
  and let $(g,D)$ be its continuation, then near the point $1$ it is
  easy to show that \begin{equation} g(z)=log(z)+2\pi i. \end{equation}
  Thus $g$ differs from $f_0$ by a constant, and is not equal to $f_0$
  near $z_0=1$.


I can't get the expression of $g$ that Lang says "It's easy to show". I tried constructing explicitly a sequence $(f_0,D_0),...,(g,D_0)$ which is the analytic continuation of $(f_0,D_0)$ but I had no success.
Essentially what is happening is that you have ended up at a different point on the associated Riemann Surface, which can be visualized by a spiral.

In more concrete terms, consider
$$ \log(e^{i\alpha})\quad \text{ for } \alpha \in [0,2\pi)$$
Which is the values we get for the logarithm as we travel around the unit circle. If we decide to follow the natural convention and have $\log(e^{i\cdot0}) = \log(1) =0,$ then we can start considering the rotated values, and note that $\log(e^{i(2\pi-\epsilon)}) = i(2\pi-\epsilon) \to 2\pi i$ as we let $\epsilon \to 0.$
Essentially what is happening is that you have ended up at a different point on the associated Riemann Surface, which can be visualized by a spiral.

In more concrete terms, consider
$$ \log(e^{i\alpha})\quad \text{ for } \alpha \in [0,2\pi)$$
Which is the values we get for the logarithm as we travel around the unit circle. If we decide to follow the natural convention and have $\log(e^{i\cdot0}) = \log(1) =0,$ then we can start considering the rotated values, and note that $\log(e^{i(2\pi-\epsilon)}) = i(2\pi-\epsilon) \to 2\pi i$ as we let $\epsilon \to 0.$