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Branch cuts in complex analysis 
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#1
Oct312, 01:46 AM

P: 57

1. The problem statement, all variables and given/known data
Given that the standard square root sqrt(anything) has a branch cut from (inf,0), find the branch cuts of the following: z+sqrt(z^21) z+isqrt(1z^2) z+sqrt(z+1)sqrt(z1) 2. Relevant equations 3. The attempt at a solution I understand what branch cuts do (multivalue functions > single value), I am just having trouble since all three of those expressions can be reduced to the same expression  I don't see why they would need different cuts? I can't really find any examples online and my book only has less than one paragraph explaining branch cuts (let alone examples). 


#2
Oct312, 03:27 AM

P: 939




#3
Oct312, 10:43 AM

P: 57

2+sqrt(2^21) = 2+sqrt(3) 2+sqrt((2)^21) = 2+sqrt(3) z+isqrt(1z^2): 2+isqrt(12^2) = 2+isqrt(3) = 2+sqrt(3) 2+isqrt(1(2)^2) = 2+isqrt(3) = 2+sqrt(3) z+sqrt(z+1)sqrt(z1): 2+sqrt(2+1)sqrt(21) = 2+sqrt(3)sqrt(1) = 2+sqrt(3) 2+sqrt(2+1)sqrt(21) = 2+sqrt(1)sqrt(3) = 2+sqrt(3) am I missing something? 


#4
Oct312, 10:46 AM

P: 1,666

Branch cuts in complex analysis
[tex]f(z)=z+\sqrt{z^21}[/tex] and I know [itex]G(z)=\sqrt{z^21}[/itex] is not a polynomial and that the quantity [itex]z^21=0[/itex] when [itex]z=\pm 1[/itex]. Therefore, from what I said above, the branch points are infinity and [itex]\pm 1[/itex]. Thus I can make branchcuts in several ways, from 1 to infinity AND 1 to infinity, or a branchcut from 1 to 1 or I can even make one from 1, through 1 and through to infinity. Now in this particular example, the standard way of making branchcuts would be two ways: (1) make a cut from 1 to 1 (2) make two cuts: one from infty to 1 and another one from 1 to infty Now how about this: even though that's not exactly what you might want, try and use what I said to devise branchcuts for the other ones. 


#5
Oct312, 01:09 PM

P: 939




#6
Oct312, 01:40 PM

P: 57

1) [tex]z^21=R[/tex] where R ranges from (0, inf). Solving for Z yields [tex]z = \pm \sqrt{1R}[/tex], which for R<1 gives (1,1), and for R>1 gives (i inf, +i inf) so I get branch cuts from (1,1) and the imaginary axis. 2) solving [tex]1z^2=R[/tex] yields [tex]z = \pm \sqrt{1+R}[/tex], which gives me (inf, 1) and (1, inf) for branch cuts. 3) Solving [tex]z+1=R, z1=R[/tex] gives (inf, 1) and (inf, 1), but the overlap cancels(?) to just give (1,1) as the branch cut. 


#7
Oct312, 03:49 PM

P: 1,666

[tex]\sqrt{1z^2}[/tex] you should look at that and immediately conclude the branch points are at 1 and 1 and the branchcuts can basically extend anyway from those points, like I said, even stairsteps going up from 1 and stairsteps going down from 1. But to make it simple, we just extend the cut from 1 to 1 (branchpoint to branchpoint) or depending on the application, from infty to 1 along the real axis is one cut and 1 to +infty on the real axis as another cut with infinity being another branchpoint. However, I'm not really sure what your particular question is asking. 


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