Branch cuts of $\frac{1}{\sqrt{z^2 + m^2}}$ I am reading up about Quantum Field theory and the integral of the following function pops up: $$\frac{1}{\sqrt{z^2 + m^2}}$$The details are actually explained in this question.
The book says that there are two branch cuts, $[-\infty, -im]$ and $[im, +\infty]$.
Now:
1) Since I can write $\frac{1}{\sqrt{z^2 + m^2}} = \frac{1}{\sqrt{z + im}\sqrt{z-im}}$, which branch cut corresponds to which $\sqrt{z\pm im}$ ?
2) Is the branch cut of $\sqrt{z - im}$, $[-\infty, im]$ or $[im, +\infty]$? 
3) If it were possible to choose the branch cuts for $\sqrt{z - im}$ to be $[-\infty, im]$ , and for $\sqrt{z + im}$ to be $[-im, +\infty]$, how do I treat the interval $[-im, im]$?
 A: I didn't look at the link post. I am going off the information giving that the function is $\frac{1}{\sqrt{z^2+m^2}}$.

I don't see why infinite branch cuts are used when we can use a single finite branch cut. If we let $z_1 = z - im = r_1e^{i\theta_1}$ and $z_2 = z + im=r_2e^{i\theta_2}$, we have
$$
\frac{1}{\sqrt{z^2+m^2}} = \frac{1}{\sqrt{r_1r_2}\exp(i(\theta_1+\theta_2)/2)}
$$
If we wind around each branch point separately, the function is multi-valued since we have
$$
\frac{1}{\sqrt{r_1r_2}\exp(i(\theta_1+\theta_2+2\pi)/2)} = -\frac{1}{\sqrt{r_1r_2}\exp(i(\theta_1+\theta_2)/2)}
$$
If we wind around both, we have
$$
\frac{1}{\sqrt{r_1r_2}\exp(i(\theta_1+\theta_2+4\pi)/2)} = \frac{1}{\sqrt{r_1r_2}\exp(i(\theta_1+\theta_2)/2)}
$$
so the function is single valued. Therefore, we can define the branch cut from $[-im, im]$
A: A branch of the square root is only defined in a branch of the logarithm, i.e., in a simply connected region of $\dot{\mathbb{C}}$. So in fact you have an infinite amount of branches: any ray from either $im$ or $-im$ to infinity defines one.
In particular, observe that if $f_{\pm}(z)=z\pm im$, then
$$
f_+(-\infty,-im] = (-\infty,0]\quad\text{and}\quad f_-[im,\infty) = [0,\infty),
$$
(these intervals are imaginary!) so we can define two different branches of the logarithm (hence two branches of the square root), one in $f_+\left(\mathbb{C}\backslash (-\infty,-im]\right) = \mathbb{C}\backslash(-\infty,0]$ and the other in $f_-\left(\mathbb{C}\backslash[im,\infty)\right) = \mathbb{C}\backslash[0,\infty)$. In order to be able to define a common branch of the logarithm you need a branch that works in both regions (the principal branch works perfectly in this case). 
Also observe that in this case the function $f:=\sqrt{f_-(z)f_+(z)}$ will be defined in:
$$
\mathbb{C}\backslash(-\infty,-im]\cap\mathbb{C}\backslash[im,\infty).
$$
This should've already answered (1) and (2). 
For (3) observe that the points in the interval are just as any other point in the plane:
$$
\begin{align}
f_-(-im,im) = (-2im,0), \\
f_+(-im,im) = (0,2im).
\end{align}
$$
(these are imaginary intervals!) Which are contained in the region where the square root is defined.
