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Complex Square Root Analyticity

  • Thread starter ferret123
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  • #1
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Homework Statement


Let f(z) denote the multivalued function [itex](z^{2} − 1)^{1/2}[/itex]
.
Define a branch of f(z) which is analytic in the interior of the unit disk |z| < 1





2. The attempt at a solution
Having a bit of trouble getting started.
I have rewritten f(z) as [itex]((z-1)(z+1))^{1/2}[/itex] as per a hint in the question.
Giving me branch points at z=1 and z=-1 and a branch cut between.
I could rewrite each part in terms of polar representation but there I get stuck on defining which argument to take for my branch.
Any help would be much appreciated.
 

Answers and Replies

  • #2
Ray Vickson
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Homework Statement


Let f(z) denote the multivalued function [itex](z^{2} − 1)^{1/2}[/itex]
.
Define a branch of f(z) which is analytic in the interior of the unit disk |z| < 1





2. The attempt at a solution
Having a bit of trouble getting started.
I have rewritten f(z) as [itex]((z-1)(z+1))^{1/2}[/itex] as per a hint in the question.
Giving me branch points at z=1 and z=-1 and a branch cut between.
I could rewrite each part in terms of polar representation but there I get stuck on defining which argument to take for my branch.
Any help would be much appreciated.
Who is forcing you to take the branch cut along the real segment [-1,1]? If you do that you cannot make your f(z) analytic in the unit disk about z = 0, at least not easily.
 
Last edited:
  • #3
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No one is forcing me to, I suppose I could alternatively take branch cuts from -1 to -infinity and from 1 to infinity. I'm still having difficulty making the next step, does my choice of branch cut like this mean I have analyticity everywhere but on the cuts or have I misunderstood them?
 
  • #4
Dick
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No one is forcing me to, I suppose I could alternatively take branch cuts from -1 to -infinity and from 1 to infinity. I'm still having difficulty making the next step, does my choice of branch cut like this mean I have analyticity everywhere but on the cuts or have I misunderstood them?
Yes, use the branch cut (-infinity,-1] for ##(z+1)^{1/2}## and [1,infinity) for ##(z-1)^{1/2}##. The function of the branch cuts is to make sure every path that circles a branch point will cross a branch cut.
 
  • #5
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So if I take my branch point at z=-1 and express ##(z+1)^{1/2}## in polar form I would let it's argument vary from (-pi, pi] so it starts and ends at the branch cut then do a similar thing around the other branch point except varying [0, 2pi)?
 
  • #6
Dick
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So if I take my branch point at z=-1 and express ##(z+1)^{1/2}## in polar form I would let it's argument vary from (-pi, pi] so it starts and ends at the branch cut then do a similar thing around the other branch point except varying [0, 2pi)?
That sounds roughly correct. Except the intervals aren't going to include the end points pi and 0, right?
 
  • #7
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Ok because then it would be crossing the branch cut? So if I express each of the parts as mentioned there that will give me an analytic branch on the interior of the unit disk?
 
  • #8
Dick
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Ok because then it would be crossing the branch cut? So if I express each of the parts as mentioned there that will give me an analytic branch on the interior of the unit disk?
Well, yes. The only places the product of the functions won't be analytic is along the branch cuts. None of those places are in the interior of the unit disk, are they?
 
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  • #9
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No so the branch is analytic there, thanks. The next part of the question asks for analyticity on the exterior of the disk with the hint ##z^{2} - 1 = z^{2}(1-z^{-2})##. I assume I approach this in a similar way except a branch cut from 1 to -1 now makes more sense?
 
  • #10
Dick
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No so the branch is analytic there, thanks. The next part of the question asks for analyticity on the exterior of the disk with the hint ##z^{2} - 1 = z^{2}(1-z^{-2})##. I assume I approach this in a similar way except a branch cut from 1 to -1 now makes more sense?
Sounds like a good strategy.
 
  • #11
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Ok thanks for all help and clarification
 
  • #12
CAF123
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That sounds roughly correct. Except the intervals aren't going to include the end points pi and 0, right?
Could you (or someone) explain why the intervals for the branch cuts don't include the endpoints? A branch cut for ##(z+1)^{1/2}## is the line extending from -1 to infinity. So ##(z+1)^{1/2}## is discontinuous across this line. I thought by leaving one of the endpoints open, it would not be a problem since we guarantee discontinuity.

Also, why is that these two branch cuts necessarily allow for analyticity in the interior of the unit disk?

Thanks.
 

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