Finding Branch Values I have a function $f(z) = (\frac{8}{7}z^3-\frac{64}{7})^\frac{1}{3}$ which I have found to have branch points at $z_1 = 2$, $z_2 = 2\exp(\frac{2i\pi}{3})$, and $z_3 = 2\exp(\frac{-2i\pi}{3})$. The part I am stuck on I am suppose to consider branch cuts from $z_1$ to $z_2$. Then, let $F(z)$ denote the branch when Im[$f(10)$] = 0. What is the value of $F(1)$. I'm confused on how exactly I find $F(z)$.  
 A: Before we tackle this problem, consider the simpler case $g(z)=z^{1/3}$, with branch points at 0 and $\infty$.  Suppose Im[$g(1)$] = 0.  For each of the possible branch cuts joining 0 and $\infty$, what is $g(-1)$?  Clearly $g(-1)$ has to be one of the 3 cube roots of -1.  The three cube roots of -1 are -1, $e^{-\pi i/3}$, and $e^{\pi i/3}$. So which one is it?
The main reason that branch cuts are necessary for $g(z) = z^{1/3}$ is because the arg function $\theta$ cannot be defined continuously for all nonzero $z$.  If we could define $\theta$ continuously, then we could take cube roots continuously by using $\theta/3$.  The reason $\theta$ can't be defined continuously is because if, say, the positive real axis is assigned $\theta = 0$, then by the time you wander all the way around a circle centered at 0, you find that the same point wants to be assigned either $\theta = 2\pi$ or $-2\pi$, depending on which direction you wander.  This is explained in more detail here.
By taking a branch cut, we remove that curve from consideration and try to define $\theta$ continuously on the rest of the complex plane, and then divide it by 3 to take a cube root (i.e., define a branch of $g(z)$).  This can be done.  Suppose we take a branch cut going from 0 to $\infty$, upwards along the imaginary axis.  We are given that Im[$g(1)$] = 0 (i.e. $g(1)=1$), and asked to find $g(-1)$.  Since Im[$g(1)$] = 0, we should choose $\theta = 0$ along the positive real axis, so that $\theta/3 = 0$ agrees with the arg of the given value $g(1)=1$.  Now we need to find a path from 1 to -1 which avoids the branch cut.  This can be any curve joining 1 to -1 that swings downwards into the lower half plane.  Because we swing downwards, by the time we get to -1, $\theta$ must be $-\pi$ (not $\pi$).  Then $\theta/3$ = $-\pi/3$.  So the cube root of -1 we want must have its argument equal to $-\pi/3$, so $g(-1) = e^{-\pi i/3}$.
Similarly, if we were to choose a branch cut going downwards from 0 to $\infty$, then $g(-1) = e^{\pi i/3}$ instead.
Now back to our original problem with $f(z)$.  You choose a branch cut between $z_1$ and $z_2$, and find a path between 10 (where you are given a value) to 1 (where you need to find the value).  This path must avoid the branch cut.  Near each of the branch points, you have a similar situation as in the $g(z)$ example above.  I'll let you work out the details.
