Branch Cuts in the Complex Plane

metgt4

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

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Dick

Homework Helper
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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