Complete analytic function for $\sqrt{1+\sqrt{z}}$.

Question

Let $U=\mathbb{C}\setminus(-\infty,0]$.

There is an analytic function $\phi$ on $U$, such that $\phi(z)^2=z$. $\phi$ is the inverse to $z\mapsto z^2$ on the right half plane. The image of $U$ under $z\mapsto 1+\phi(z)$ is inside $U$ so we can define $f: U\rightarrow \mathbb{C}$ such that $z \mapsto \phi(1+\phi(z))$.

The complete analytic function for $(U,f)$ is the set of all function elements $(V,g)$ where $V\subset \mathbb{C}$ is connected and open and $g:V\rightarrow \mathbb{C}$ is analytic, such that $(V,g)$ is an analytic continuation of $(U,f)$. I need to find all the germs in the complete analytic function at $1/2$ and at $1$.

Here is my attempt: Let $(V,g)$ be an analytic continuation of $(U,f)$ such that $1\in V$.

Since $(f(z)^2-1)^2=z$ on $U$, then $(V,(g^2-1)^2)$ is an analytic continuation of $(U,\mathrm {id}_U)$. Then since the identity map is an entire function, $(g^2-1)^2=\mathrm{id}_V$. This means that $g(1)=\pm \sqrt2, 0$. Since $h(z)=(z^2-1)^2$ is a left inverse to $g$, $h'(g(1))g'(1)=1$. $h'(0)=0$ so $g(1)\neq 0$. I am left with the possibilities $g(1)=\pm \sqrt2$. Since $h$ is an analytic left inverse to $g$, local behavior of $h$ at $g(1)$ determines $g$ near $1$, so there is at most one germ corresponding to each of $g(1)=\pm \sqrt2$. I know already that $f(1)=\sqrt 2$ so it suffices to prove that $(U,-f)$ is an analytic continuation of $(U,f)$. I feel this is true after drawing some loops in the complex plane.

Similarly for $1/2$. Let $(V,g)$ be an analytic continuation of $(U,f)$ such that $1/2\in V$. Then $f(1/2)=\pm\sqrt{1\pm\sqrt{1/2}}$. I want to show that $\pm \phi\circ(1\pm \phi)$ (well defined near $1/2$) on some small domain around $1/2$ are all analytic continuations of $f$.

Is there a neat and rigorous way to see these? Can anyone offer some ideas?

Answer 1

Some ideas:

1. You have the composition of $f(z)=\sqrt{1+z}$ and $g(z)=\sqrt{z}$.
2. $g$ has two germs at any nonzero point. It is not analytic at $0$.
3. $f$ has two germs at any point other than $-1$. It is not analytic at $-1$.
4. The composition $f\circ g$ has four germs at $1/2$, whose values at $1/2$ correspond to the choices of $\pm $ in $\pm \sqrt{1\pm \sqrt{1/2}}$
5. The composition $f\circ g$ has two germs at $1$, whose values at $1$ correspond to the choices of $\pm $ in $\pm \sqrt{1+ \sqrt{1}}$. The germ of $g$ that has $g(1)=-1$ cannot be inserted in $f$, per item 2.