# Define a branch for $\sqrt{1+\sqrt{z}}$ and show it is analytic.

Define a branch for $$\sqrt{1+\sqrt{z}}$$ and show it is analytic.

I defined a branch $$(-\pi, \pi)$$, and so that means the function $$\sqrt{1+\sqrt{z}}$$ is analytic on $$\mathbb{C}\setminus \left\{y=0,x\leq 0\right\}$$.

I am trying to analyze when such a function is in the deleted area. Already from $$\sqrt{z}$$, $$y$$ must be zero. It remains to consider the case when $$x\leq 0$$. Since there's an added $$+1$$ inside the square root, does this change $$x\leq 0$$ to $$x\leq -1$$? And so the analytic domain is $$\mathbb{C}\setminus \left\{y=0,x\leq -1\right\}$$?

If $$1+\sqrt z= -t$$ with $$t \geq 0$$ then $$z=(1-t)^{2}$$. The range of $$(1-t)^{2}$$ on $$[0,\infty)$$ is $$[0,\infty)$$ so $$z \in [0,\infty)$$ which is not true. Hence $$1+\sqrt z$$ does not lie on the negative real axis when $$z$$ does not lie on the negative real axis. Hence $$\sqrt {1+\sqrt z }$$ is well defined and analytic on the complex plane with $$(-\infty, 0]$$ removed.
$${w = \sqrt{1 + \sqrt{z}},( z \; \epsilon \; C) \mapsto circle \; |w|\measuredangle\phi,\ by \ Moivre \ theorem \ to \ determine \ the\ complex\ roots \ it \ follows \ that: \ w^n \ = \ r^n \ ( cos(\ n \phi )\ + sin(\ n \phi) ) \ let's \; define \; this \; circle \; as \; f(z).}$$ $${ A \;positive \ branch \; is \; defined \; in \ the \; domain \; 0\le\phi< 2\pi \; , ( or \; restricting \; z,) \Rightarrow hence \;} z0 \ne \ z1 \; is \; fulfilled \; that\; f(z0) \; \ne \; f(z1), \; hence \; f(z) \; is \; analityc.$$ }