Properties of Complex Function $f(z)=e^{-\frac{1}{z^{1/3}}}$

Question

This post will be about a part of an example from my complex analysis book.

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Problem:

They claim that there exist a function $J_+(z)$ holomorphic in the upper half plane $\operatorname{im}z>0$, smooth on the real line, that does not extend holomorphically in any neighborhood of $0$ in $\mathbb{C}$.

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Construction of Function:

Their candidate is constructed by first letting $z_+^{1/3}$ be the cube root of $z$ determined by $$z_+^{1/3}=\operatorname{exp}\Big(\frac{1}{3} (\log|z|+i\arg z \Big),\quad -\pi/2<\arg z<3\pi/2.$$ They then claim that

$$J_+(z)=\operatorname{exp}(-1/z_+^{1/3})$$

satisfies the properties.

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Questions (Technicalities):

1. What is the point of defining the cube root?

Isn't $\operatorname{Log}z=\log |z|+i\arg z$ and $$e^{\frac{1}{3}\operatorname{Log}z}=z^{\frac{1}{3}}.$$ Couldn't we just have taken $$J_+(z)=e^{-\frac{1}{z^{1/3}}}?$$ Or do they perhaps do the cube-root thing to clarify something, or perhaps you need it for some technical reasons?

2. Why did they choose $-\pi/2<\arg z<3\pi/2$? Would there be any interval considered a bad choice? If so, which one?

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Questions (Proof of properties):

1. First I will try to prove that the function, $J_+(z)$, is holomorphic in the upper half-plane. Is it enough to prove that $f(z)=e^{z}$ and $-\frac{1}{z^{1/3}}$ are holomorphic in the upper half-plane. Since then their composition $$f\Big (-\frac{1}{z_+^{1/3}}\Big)=J_+(z)$$ is holomorphic in the upper half-plane?
2. Next thing to prove is that $J_+(z)$ is smooth on the real line.

Does this mean I should differentiate $J_+(z)$ as a real function, that is, fix $y=0$ if $z=x+iy$? If that's the case, couldn't I just mimic the first example given on this wikipedia article:

https://en.wikipedia.org/wiki/Non-analytic_smooth_function?

If not, how would i prove this statement?

3. Furthermore, to prove that $J_+(z)$ does not continue holomorphically past the origin, can I do the power series proof as they do in wikipedia link, or do you know any better way to prove this?

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There are already a lot of questions to be answered, I apologize for that. However, I do have two bonus questions (they are not prio one to be answered).

It doesn't feel like I have used the property of the third root anywhere. Wouldn't for that matter $$e^{-1/\sqrt{z}}$$ also work? Or why not more generally $$e^{-1/z^{1/n}}$$ for $n\in \{1,2,...\}$? Probably there is something important going on with the cube root, but I just can't see it (I just thought it is quite odd to specifically give the cube root as an example of a function).

I really hope I can get some help to understand this example. :)