# If $f(0) = 0$, $f$ injective, then there is single-valued branch of $\sqrt{f(z^2)}$

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?

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 $$\sqrt {f(z^{2})}$$