What are the Singularities of f(z) = log(1+z^1/2)?

  • Thread starter Thread starter stephenkeiths
  • Start date Start date
  • Tags Tags
    Singularities
stephenkeiths
Messages
53
Reaction score
0

Homework Statement


Find all the singularities of

f(z)=log(1+z^{\frac{1}{2}})

Homework Equations


Well I need to expand this. Find if it has removable singularities, poles, essential singularities, or non-isolated singularities. The problem is the branches. I know z^{\frac{1}{2}}
has branch points at 0 and ∞ and logz has branch points at 0 and ∞. So if we choose the '-' branch of z^{\frac{1}{2}}, then z=1 is also a branch point.

I'm just having trouble since I can't expand this function for all the different branches.
 
Physics news on Phys.org
I figured this out, should someone ever stumble upon this and be curious.

The singularities arise only due to the branch cuts! So any point on a branch but is a non-isolated singularity. However, the branch cut is somewhat arbitrary, so long as the cut ends at branch points. So the only points that MUST be singularities are the branch points. And I fully classified the branch points above.

Conclusion: there are 2 branch point singularities (that is what my book calls them) if you take the '+' branch of z^1/2, namely 0 and infinity. There are 3 branch point singularities if you take the '-' branch, namely 0, 1, and infinity.
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
Back
Top