How to find branch points I'm solving a set of exercises to understand how to find branch points and branch lines, but I'm having trouble with the more difficult('ish) ones.
What I usually do in more simple exercises, is that first I see where the given function $f(z)$ is not defined, and work with these points, let say $z_0,z_1,...$. Then I consider the argument of $f(z_i)$, and turn around it $2\pi$ to see if there's any change, until now, in all my exercises this had happened. However, I was told by a friend, that one not only you have to turn around these points, but also turn around all of them, I don't understand this nor I see how do you do this.
My struggles started with this function $f(z)=\ln(z-z^2)$, we now that $\ln$ has issues when $z-z^2=0$, i.e., $z(1-z)=0$, if $z_0=0$ or $z_1=1$. Now, how do I show that these two points are branch points? How do I turn around both?
 A: The function $\log(z-z^2)$ can be defined by integrating $ \frac{1-2z}{z-z^{2}} $ on the complex plane from $\zeta = \frac{1}{2} + \frac{i\sqrt{3}}{2}$ (where $\log(\zeta-\zeta^{2})=0$).
Notice that $ \frac{1-2z}{z-z^{2}} = \frac{1-2z}{z(1-z)}$ has simple poles at $z=0$ and $z=1$, both with residue $1$.
If we integrate counterclockwise around a closed loop with either the point $z=0$ or the point $z=1$ (but not both) in its interior, the value of $\log(\zeta-\zeta^{2})$ jumps by $2 \pi i $.  This shows that $z=0$ and $z=1$ are branch points of $\log(z-z^{2})$.
Also notice that if we integrate counterclockwise around a closed loop with both the points $z=0$ and $z=1$ in its interior, the value of $\log(\zeta-\zeta^{2})$ jumps by $4 \pi i$.
So that $\log(z-z^{2})$ is well-defined on the complex plane, we need to prevent the above situations from occurring.
One possible option, which coincides with using the principal branch of the logarithm, is to place branch cuts on the half-lines $(-\infty,0]$ and $[1,\infty)$.
A: Assuming you're using the standard branch of the logarithm (you define $\ln$ on $\mathbb{C} \backslash \mathbb{R}_{\leq 0}$), what you seem to get is two branch lines, not points. You won't only have problems when the argument of $\ln$ is $0$ but when it is negative on the real line. This happens when $z(1-z) \leq 0$, so exactly on the two rays $\{\Re z \leq 0\}$ and $\{\Re z \geq 1\}$ where $\Re z$ means the real part of $z$. Now I don't know your definition of a branch point or branch line in this situation, but I guess these 2 will be branch lines?
