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I need to show that if $f$ is analytic and injective in a neighborhood of $0$ and $f(0) = 0$, then there is a single-valued branch of $\sqrt{f(z^2)}$ in a possible smaller neighborhood of $0$.

The hypothesis implies that the zero at $0$ is of order $1$ (by open mapping Theorem). Thus, we can write $f(z) = zg(z)$ in a neighborhood, $D_r(0)$, of $0$ where $g(0) \neq 0$. Thus, $f(z^2) = z^2g(z^2)$. Since $g$ is nonzero in a $D_r(0)$ and $D_r(0)$ is simply connected, we can define single valued branch of $\sqrt{g(z^2)}$. However, how can I define a branch of $\sqrt{z^2}$? Naively, I want to define $\sqrt{z^2} = z$ which is true in any domain. Can I say this?

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Let $h(z)= z \sqrt {g(z^{2})}$. Then $h$ is single valued, analytic and $h(z)^{2}=z^{2}g(z^{2})=f(z^{2})$. Hence $h$ is an analytic branch of $\sqrt {f(z^{2})}$

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