I'll try (i) for $\displaystyle \alpha=1/2$:
You can let $\displaystyle z=1+se^{i\phi}$ and $\displaystyle z=re^{i\theta}$ then $\displaystyle \sqrt{\frac{re^{i\theta}}{se^{i\phi}}}=\left(\frac {r}{s}\right)^{1/2}e^{i/2(\theta-\phi+2k\pi)}$ then $\displaystyle \arg f=\frac{\theta-\phi}{2}+2k\pi$. Now consider (i) at various points around the circle $\displaystyle 1/2 e^{it}$. At 1/2, $\displaystyle \theta=0$ and $\displaystyle \phi=\pi$ so $\displaystyle \arg f(1/2)=-\pi/2$. Now start going around that circle and show how the term $\displaystyle \frac{\theta-\phi}{2}$ changes smoothly between the values of $\displaystyle -\pi/2$ and $\displaystyle 3\pi/2$ except when you cross the line segment between zero and one and therefore the function $\displaystyle f(z)=\sqrt{z/(z-1)}$ has a single-valued analytic component over this circle except for the line segment (branch-cut) between zero and one.