Branch cut of $\sqrt{z}$ In my complex analysis book, the author defines the $\sqrt{z}$ on the slit plane $\mathbb{C}\setminus (-\infty,0]$. I understand this is done because $z^2$ is not injective on the entire complex plane, so we restrict the domain of $z^2$ to $(\frac{-\pi}{2},\frac{\pi}{2})$, and then $z^2$ maps bijectively on $\mathbb{C}\setminus(-\infty,0].$ This particular restriction of the domain to $(\frac{-\pi}{2},\frac{\pi}{2})$ give rise to the principle root function. My question is, why can't we include either the positive or negative real axis in this domain? I understand why $[\frac{-\pi}{2},\frac{\pi}{2}]$ wouldn't work, as $(-\infty,0)$ would be mapped to twice, but it seems to me that $z^2:[\frac{-\pi}{2},\frac{\pi}{2})\to\mathbb{C}$ would define a bijection and allow us to define $\sqrt{z}$ as single-valued. So why don't we do this?

The domain of $w=f(z)=z^2$ is not restricted to $(\frac{-\pi}{2},\frac{\pi}{2})$ to obtain an inverse. Rather, the interval $[0,\pi ]$ (or $[\pi ,2\pi ]$ is obtained as the image of a bijection from the slit $w$ plane. This map is then a branch of the inverse. 
The procedure is as follows: 
it's no harder to look at the more general case so let $w=f(z)=z^{n}, \ \text {for }n\in \mathbb N$. Then $f$ maps the sector $0\leq \theta \leq2\pi/n$ onto $\mathbb C$. 
To every point on the positive real axis of the $w$ plane there corresponds one point on each of the two rays $\theta =0$ and $\theta =2\pi/n$ in the $z$ plane.  Therefore, except for the positive real axis in the $w-$ plane, the map is bijective. 
So we can get a bijection if we "cut" the positive real axis of $w$ plane so that after the cut, the upper edge corresponds to the positive real axis in the $z$ plane and the lower edge corresponds to $\theta =2\pi/n$ in the $z$ plane. 
We now simply define $z=w^{1/n}$ to be the inverse of this map. It maps the slit $w$ plane bijectively onto the sector $0\leq \theta \leq 2\pi/n$.
It is easy to see that there are sectors $\frac{2\pi k}{n}\leq \theta \leq \frac{2\pi (k+1)}{n}, \quad k=0,1,\cdots n-1$ in the $z$ plane onto which are mapped bijectively, copies of the slit $w$ plane and so we get by this process $n$ "branches" all of which are inverses of $w=z^n$. They map slit copies of $\mathbb C$ onto sectors in the $z$ plane. 
A moment's reflection shows that we can take $\textit {any}$ sector that sweeps out an angle $2\pi /n$ in the $z$ plane and the foregoing analysis goes through with very minor modifications. 
A nice way to tie all this up is to now glue all these copies together so as to define a injection from the glued copies back to the entire $z$ plane. i.e. consider the Riemann Surface for $w=z^n$. 
