I've been trying to arrive at a branch cut that works for this function.
Firstly, I find the branch points which are, if I'm not mistaken, $z_1=0$ and $z_2=1$.
I then tried encircling them to see whether $f(z)$ changed its value when I completed a $2π$ spin around each branch point and around both simultaneously. Doing this, I find that $f(z)$ does change its value in every kind of curve; around only $z_1$, around only $z_2$, and around both. So, since I'm working with $\operatorname{Arg}(z)∈(-π,π]$, my branch cut will be a line across the x axis for $x\leq1$.
That would have completed the problem but then I tried doing it a different way. I went this way:
Since the branch cut for $\operatorname{Log}(z)$ is the negative real axis, the points $(x,y)$ that form this branch cut will be those such that
$$\operatorname{Re}(g(z))\leq0$$
and
$$\operatorname{Im}(g(z))=0$$
with
$$g(z)=\frac{1}{z(z-1)}.$$
Operating with both conditions, I get two possible branch cuts
$$(x,y)/x∈[0,1], y=0$$
and
$$(x,y)/x=\frac{1}{2}, y∈\left[-\frac{1}{4},∞\right]$$
I immediately discard the second option since it doesn't even go through the branch points and I'm left with the first option: connecting the branch points. But this goes against what I said earlier about how making a 2$\pi$ turn around both points is supposed to change $f$'s value, but if I'm choosing this branch cut, $f$'s value shouldn't be changing with such a spin.
My guess is that I'm doing the encircling a point part wrong. But I'm just starting with multivalued functions so I could have made mistakes anywhere.
Any help regarding how to analytically find branch cuts/points for $\operatorname{Log}(g(z))$ in general and for $g(z)=\frac{1}{z(z-1)}$ in particular would be very much appreciated. Thanks in advance