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

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SUMMARY

The function f(z) = log(1 + z^{1/2}) has singularities that arise exclusively from branch cuts. The analysis reveals that there are two branch point singularities at 0 and infinity when using the '+' branch of z^{1/2}. Conversely, when the '-' branch is selected, there are three branch point singularities at 0, 1, and infinity. The classification of these singularities is crucial for understanding the behavior of the function in complex analysis.

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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.
 
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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.
 

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