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Branch Cuts in the Complex Plane

  • Thread starter metgt4
  • Start date
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1. Homework Statement

The function

f(z) = (1-z2)1/2

of the complex variable z is defined to be real and positive on the real axis in the range -1 < x < 1. Using cuts running along the real axis for 1 < x < infinity and -infinity < x < -1, show how f(z) is made single-valued and evaluate it on the upper and lower sides of both cuts.


3. The Attempt at a Solution

I started by expressing f(z) in terms of complex exponentials:

f(z) = (1-z2)1/2
= (e-i2pi - r1r2ei2(theta1 + theta2))1/2
= (r1r2)1/2ei(theta1 + theta2 - pi)

But I don't understand how to evaluate this on both sides of the cuts. I've searched the internet quite a bit trying to find something to help explain it to me but couldn't find anything so I've come to you guys!

Thanks for any help!
Andrew
 

Dick

Science Advisor
Homework Helper
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Remind yourself of how sqrt(z) is defined around 0 with a branch cut along the negative x axis. You define choose the range of argument theta=(-pi,pi) for z not on the negative real axis and then if z=r*e^(i*theta), define sqrt(z)=r^(1/2)*exp(i*theta/2), right? For your function you want to define sqrt(1+z) around z=(-1) and sqrt(1-z) around z=1 with their suggested cuts in an analogous way. Then multiply them together to get f(z).
 

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