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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 $\mathbb{C}\setminus \left\{y=0,x\leq -1\right\}$?

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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.

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$ {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. $ }

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  • $\begingroup$ Rectifiqué la respuesta despues de repasar el tema, no había entendido bien que se pedía. Sorry $\endgroup$ – Carlos Obiglio Mar 28 '20 at 21:34

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