# 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 $$<1$$ but $$>0$$. 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.

## 1 Answer

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 $$\epsilon \to 0.$$