Simple complex numbers: Branch points

[adwodon]

Homework Statement

f(z)=\sqrt{(z.^3+8)}

How many branches (solutions) and branch points does the funtion f(z) have?

Homework Equations

The first part of the question was working out the roots of z^3+8=0 which I found to be -2, 1+i\sqrt{3} and 1-i\sqrt{3}

The Attempt at a Solution

I would just like some clarification as to the difference between branch points and branches?
Would I be right in saying it had an infinite number of solutions (which the question says are branches)? As z=2e^i(\pi+\frac{2}{3}n\pi) where n=0 to infinity?
And that it has 3 branch points.

So if z was then square rooted you would square root each branch point and get 2 new branch points on each original point so you would have 6 total branch points, or maybe 9 (6 new ones + 3 original ones? Or am I thinking about this is completely the wrong way...

 

[Bacle]

Hope this is still of interest

A branch point for a function f(z) is a point zo so that if you wind around zo, say
with a circle, you do not go back to your initial value. As an example, take
f(z)=z^1/2, and find the value of f(z) for e^it and e^i(t+2Pi).

A branch for z^1/2 is a region of the plane where you can draw no curves with
the above property, i.e., curves where going around gives you different values.

Too construct a branch, you then remove a region of the plane containing all
problem (branch) points, so that you cannot draw any such curves. The standard
example maybe is that of Logz, where you remove the negative real axis, so that
no curve can wind around the origin--the branch point for log.

Your square root function is not entire, as it is not even single-valued. You
need to:

i)Restrict to avoid multi-valuedness

ii)Restrict the plane to avoid branch points.

You need to define a branch for each of your square root functions --define a
square root function for each root. The branch is then one that works for _all_
the functions, i.e., an intersection. The branch points will happen at the roots,
check it out.