Consider the function $\sqrt{z}$, with $z\in\mathbb{C}$.
Writing $z = re^{i\theta}$, the imaginary part of $\sqrt{z}$ can be expressed as: $$Im({\sqrt{z}) = r^{\frac{1}{2}}\sin({\frac{\theta}{2} + k\pi})}$$
with $k\in\mathbb{Z}$. For the time beind let's set $k=0$ anyway.
Now, if I choose the principal value of the function to be $-\pi<\theta<\pi$, I can cleary see that $Im(\sqrt{z})$ has a branch cut along the negative real axis because $$\sin(\pi/2) \neq \sin(-\pi/2)$$.
Now: I want the branch cut to be along the positive real axis. I would set the principal value to be $0<\theta<2\pi$, but then, along said axis, $\sin(0/2) = \sin(2\pi/2) = 0$ so there is no discontinuity...
A branch cut is a curve or a line that is introduced in order to define a branch $F$ of a multivalued funtion, in this case $f(z) = \sqrt{z}$. Now points of a branch cut are singular points of $F$. You want to be along the positive real axis, the origin and the ray $\theta = 0 $ will make up the branch cut.
Notice that the discontinuity will be if we take $\sqrt{z} = \sqrt{r}e^{i\frac{\theta}{2}} = \sqrt{r}\Big(\cos \Big(\frac{\theta}{2}\Big) + i\sin\Big(\frac{\theta}{2}\Big)\Big)$, where $0 < \theta < 2\pi$. Then
$$\lim_{\epsilon \to 0^+} \cos \Big(\frac{\epsilon}{2}\Big) \neq \lim_{\epsilon \to 0^-} \cos \Big(\frac{2\pi - \epsilon}{2}\Big).$$