You can first shift to the origin by letting u= z- z0 so your question refers to f(u)= ln(u). Now any straight line, from 0 to infinity can be taken as a branch cut! The "standard" choice is to take the negative x-axis as branch cut so that [itex]ln(z)= ln(re^{i\theta})= ln(r)+ i\theta[/itex] and [itex]-\pi< \theta< \pi[/itex]. The point is that that choice of branch cut "separates" [itex]-\pi< \theta< \pi[/itex] from [itex]\pi< \theta< 3\pi[/itex] from [itex]3\pi< \theta< 5\pi[/itex], etc.
In terms of your original z, a branch cut is a straight line from z0 to infinity and the "standard" choice is the line z= t+ iy0 where y0 is the imaginary part of z0 and t can be negative real number.