Example 10.8 (continuation of \(\log z\) around the origin). On the disc \(D_0 = \left |z - 1\right | < 1\) define the principal logarithm \[f_0(z) = \operatorname {Log} z = \sum ^{\infty }_{n = 1} \frac {(-1)^{n-1}}{n}(z - 1)^n ,\] which is analytic there and satisfies \(e^{f_0(z)} = z\) with \(f_0(1) = 0\).

Continue \(f_0\) along the unit circle \(\gamma (t) = e^{2\pi i t}\), \(t \in [0,1]\). At each stage the element is a branch of the logarithm on a disc about \(\gamma (t)\), and the value carried along the path is \[f_t\big (\gamma (t)\big ) = 2\pi i t .\] At \(t = 1\) the path has returned to its starting point \(z = 1\), but the element it has arrived with satisfies \[f_1(1) = 2\pi i \neq 0 = f_0(1).\] The continuation exists all the way round and is unique — and it does not come back to where it began.

Nothing has gone wrong. The lesson is that \[\log z = \ln \left |z\right | + i\operatorname {arg}(z)\] has no single-valued analytic continuation to the punctured plane \(\mathbb {C}\setminus \{0\}\); each circuit of the origin adds \(2\pi i\). The same computation with \(\sqrt {z}\) returns \(-\sqrt {z}\) after one loop and only closes up after two.

The obstruction is not the size of the domain but its shape: the punctured plane has a hole, and the path encircles it.