dwsmith said:
Given the following functions:
\[
f(z) = \sqrt{\frac{z}{1 - z}}
\]
The branch points are then \(z = 0\) and \(z = 1\) and the branch cut is the line from \((0, 1)\), correct?
Yes, you are correct. $f(z)$ has two sheets ($f_1$ and $f_2$) over $\Bbb C$, one being the square root with a plus sign another with the minus sign, and the ramification point of those two sheets occur precisely at $z = 0$, where $f_1(z) = f_2(z) = 0$ and at $z = 1$ where $f_1(z) = f_2(z) = \infty$. The branch cut is the straightline joining those branch points.
\[
f(z) = (z^2 - 4)^{1/3}
\]
Here the branch points are \(z = \pm 2\). So would the branch cut be \((-2, 2)\) or \((-2, 0)\) and \((0, 2)\) or something entirely different?
The function has three sheets. Each of them are ramified at $z = \pm 2$, as at those points $f_1(z) = f_2(z) = f_3(z) = 0$. The branch cut is simply $[-2, 2]$. $[-2, 0]$ or $[0, 2]$ are not really plausible candidates as $0$ is not even a branch point, so I am not sure from where you are getting them.
\[
f(z) = \ln(z - z^2) = \ln\lvert z - z^2\rvert + i(arg(z - z^2) + 2\pi k)
\]
Not sure what to say about this one.
This is a wee bit weird. The Riemannsurface of this function gives an infinite-sheeted covering over $\Bbb C$, with one logarithmic branch point at $z = 0$. (the sheets aren't quite ramified at this point, as letting a point loop around $z = 0$ returns an infinite winding number). Other one is at $z = 1$ it seems. A bit rigorously, $\displaystyle \log(z - z^2) = \int_\omega^z \frac{1-2t}{t-t^2} \, \mathrm{d}t$ and the integrand has simple poles at $t = 0, 1$ of residue $1$. You can verify by some clever looping-type argument that these are both branch points. However, looping around both of them once jumps the function by $\pm 4\pi i$ so $z = \infty$ is another branch point.
EDIT : Oops I forgot to produce the branch cut for the last one. The possible branch points are $z = 0, 1, \infty$ but the branch cut connecting all of the branch points will remove the possibility of moving around the branch points on the complex plane. So the only possible branch cut is $[0, 1] \cup \bf{\Gamma}$ for any path $\bf{\Gamma}$ on $\Bbb C\cup\{\infty\}$ joining $0$ and $\infty$.
