Branch points of a complex function

[Hill]

Homework Statement

Consider the multifunction ##f(z) = \sqrt {z - 1} \sqrt[3] {z - i}##.
Where are the branch points and what are their orders?

Relevant Equations

##e^{i \frac \theta n} = e^{i \frac {\theta + 2 \pi n} n}##

My answer: one branch point is ##1## of the order 1, another is ##i## of the order 2.
My question is, how can I be sure that these are the only branch points?

 

[andrewkirk]

Prove by construction that, at an arbitrary point in the complex plane excluding 1 and ##i##, each branch of the function ##f## has a continuous inverse. At least one branch of a function does not have a continuous inverse at a branch point. So if every branch has a continuous inverse, it cannot be a branch point.

 

[WWGD]

Prove by construction that, at an arbitrary point in the complex plane excluding 1 and ##i##, each branch of the function ##f## has a continuous inverse. At least one branch of a function does not have a continuous inverse at a branch point. So if every branch has a continuous inverse, it cannot be a branch point.

Yes, good point, branch point of product doesn't necessarily coincide with the intersection of the branch points. Could this be expressed in terms of the monodromy groups?

 

[Hill]

Prove by construction that, at an arbitrary point in the complex plane excluding 1 and ##i##, each branch of the function ##f## has a continuous inverse. At least one branch of a function does not have a continuous inverse at a branch point. So if every branch has a continuous inverse, it cannot be a branch point.

If I understand this test correctly:

Let's take a simpler function, ##f(z)=\sqrt z##. It has branch point at 0.
The inverse of this function is ##z(f)=f^2##.
Isn't ##z(f)## continuous everywhere, including the branch point ##f(0)=0##?