Branch points of global analytic functions (Ahlfors) In Ahlfors' complex analysis text, page 298 he discusses the case of a global analytic function $\mathbf{f}$ which can be continued analytically along all arcs in some punctured disk $\{0<|z|< \rho\}$. He then fixes a point $0<z_0=r<\rho$, and an initial germ of $\mathbf{f}$ over that point, and continues that germ along all integer powers of the circle $C=\{|z|=r\}$. There are are two cases:


*

*The germ obtained after $m$ windings around the circle never equals the initial one.

*There is a smallest integer $h$, such that tracing the circle $h$ times leads back to the initial germ.


Only the second case is considered further: Ahlfors then says that such a function "can be expressed as a single valued analytic function $F(\zeta)$" where $\zeta$ is related to $z$ by the equation $$z=\zeta^h .$$
So far this is only a background, my question regards page 299: On page 299 he constructs $F(\zeta)$ using the initial germ, and expands it in a Laurent series:
$$F(\zeta)=\sum_{-\infty}^\infty A_n \zeta^n. $$
He then writes:

It must be observed that this development depends on the choice of the initial germ; different choices may yield entirely different developments and, in particular, different values of $h$. Actually, even the series $(1)$ yields $h$ different developments, corresponding to the $h$ initial values of
  $z^{1/h}$. If we write $\omega = 
e^{2 \pi i/h},$ these developments are represented by $$\tag{2} f_\nu(z)=\sum_{-\infty}^\infty A_n \omega^{\nu n} z^{n/h} \; \; (\nu=0,1,\dots,h-1) $$
  When the germ $(f_\nu,z_0)$ is continued along $C$ it leads to $(f_{\nu+1},z_0)$ with the
  understanding that the subscript $h$ is identified with $0$.

I have a few questions regarding the quoted part above:


*

*What does he mean by the initial values of $z^{1/h}$?

*What does he mean by using the word "developments"?


Obviously, the Laurent development of $F(\zeta)$ holds in the entire punctured disk, but where do these "Puiseux developments" $\{f_\nu\}$ hold?


*

*How can I rigorously show that continuing $(f_\nu,z_0)$ along $C$ leads to $(f_{\nu+1},z_0)$?


Intuitively, I guess as $z$ traces the circle $C$ once, $z$ transforms by the map $z \mapsto e^{2\pi i } z $, and from there $(e^{2 \pi i} z)^{n/h}=\omega^n z^{n/h}$. However, I know that using the power rule $(z w)^r=z^r w^r$ isn't that straightforward for complex numbers...
Any help is greatly appreciated, thanks!
 A: 
What does he mean by the initial values of $z^{1/h}$?

The initial germ $\mathbf{f}$ is given by a holomorphic function $f_0$ on a small neighbourhood $U$ of $z_0$. We may assume $U$ simply connected. Then there are $h$ branches of the $h$-th root defined on $U$, let's call $p_0$ the branch with $p_0(r) \geqslant 0$, and $p_\nu(z) = \omega^{-\nu}\cdot p_0(z)$, for $\nu = 1,\dotsc,h-1$. Then $\zeta \mapsto f_0(\zeta^h)$ is analytic on each of the $V_\nu = p_\nu(U) = \omega^{-\nu}\cdot V_0$, and we thus get $h$ function elements (in $h$ different points of the annulus $0 < \lvert\zeta\rvert < \rho^{1/h}$) $(g_\nu,V_\nu)$ with the property $g_\nu \circ p_\nu = f_0$. Each of the $g_\nu$ has an analytic continuation to the entire annulus $0 < \lvert\zeta\rvert < \rho^{1/h}$, which we also denote by $g_\nu$.
The "initial values of $z^{1/h}$" are the $p_\nu(z)$ on $U$ from which the analytic continuation is started. Perhaps it would be better to speak of the branches of the $h$-th root here.
Each of the $g_\nu$ has a Laurent development
$$g_\nu(\zeta) = \sum_{-\infty}^\infty A_n^{(\nu)} \zeta^n$$
and writing $\zeta = z^{1/h}$ in that yields a development of the form $(2)$. Since $g_\nu(\zeta) = g_0(\omega^\nu\cdot\zeta)$, we have $A_n^{(\nu)} = A_n^{(0)}\cdot \omega^{\nu\cdot n}$, which gives exactly $(2)$.

What does he mean by using the word "developments"?

Representations of $f$ as a series, here a "fractional power" series. "Expansions" could have been used too.

Obviously, the Laurent development of $F(\zeta)$ holds in the entire punctured disk, but where do these "Puiseux developments" $\{f_\nu\}$ hold?

On any subset of the annulus $0 < \lvert z\rvert < \rho$ on which a branch of the $h$-th root is defined. Usually, one would remove one radius from the annulus to get as large as possible a domain on which the expansion holds.

How can I rigorously show that continuing $(f_\nu,z_0)$ along $C$ leads to $(f_{\nu+1},z_0)$?

By continuing $g_\nu$ along the (positively oriented) arc from $p_\nu(z_0)$ to $p_{\nu+1}(z_0)$. That arc is mapped by $\zeta\mapsto \zeta^h$ to $C$, and since it is locally invertible, analytic continuation along the one corresponds to analytic continuation along the other via composition with an appropriate branch of the $h$-th root.
