# branch cuts in complex analysis

by csnsc14320
Tags: analysis, branch, complex, cuts
 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^2-1) z+isqrt(1-z^2) z+sqrt(z+1)sqrt(z-1) 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).
P: 927
 Quote by csnsc14320 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
No they can't! That is precisely the point. Perhaps you should calculate what you get when you plug in, say, z=±2 into those expressions.
P: 57
 Quote by clamtrox No they can't! That is precisely the point. Perhaps you should calculate what you get when you plug in, say, z=±2 into those expressions.
z+sqrt(z^2-1):
2+sqrt(2^2-1) = 2+sqrt(3)
-2+sqrt((-2)^2-1) = -2+sqrt(3)

z+isqrt(1-z^2):
2+isqrt(1-2^2) = 2+isqrt(-3) = 2+sqrt(3)
-2+isqrt(1-(-2)^2) = -2+isqrt(-3) = -2+sqrt(3)

z+sqrt(z+1)sqrt(z-1):
2+sqrt(2+1)sqrt(2-1) = 2+sqrt(3)sqrt(1) = 2+sqrt(3)
-2+sqrt(-2+1)sqrt(-2-1) = 2+sqrt(-1)sqrt(-3) = -2+sqrt(3)

am I missing something?

P: 1,639

## branch cuts in complex analysis

 Quote by csnsc14320 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^2-1) z+isqrt(1-z^2) z+sqrt(z+1)sqrt(z-1) .
It's tough. Really tough as you indicated by the lack of good information in your text about it. How about this, suppose I have a function that ISN"t a polynomial. I mean trigs, logs, other non-polynomials. Call that function G. And I tell you wherever G is zero or infinity, there is a branch-point and to make a branch-cut, I define a line from one branch-point to another. So Take:

$$f(z)=z+\sqrt{z^2-1}$$

and I know $G(z)=\sqrt{z^2-1}$ is not a polynomial and that the quantity $z^2-1=0$ when $z=\pm 1$. Therefore, from what I said above, the branch points are infinity and $\pm 1$. Thus I can make branch-cuts in several ways, from -1 to infinity AND 1 to infinity, or a branch-cut 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 branch-cuts 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 branch-cuts for the other ones.
P: 927
 Quote by csnsc14320 2+isqrt(1-2^2) = 2+isqrt(-3) = 2+sqrt(3)
Are you using √(-1) = i or √(-1) = -i?
P: 57
 Quote by jackmell Now how about this: even though that's not exactly what you might want, try and use what I said to devise branch-cuts for the other ones.
OK, so if I try to map just the square roots to the range (-inf, 0) I get:

1) $$z^2-1=-R$$ where R ranges from (0, inf).
Solving for Z yields $$z = \pm \sqrt{1-R}$$, 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 $$1-z^2=-R$$ yields $$z = \pm \sqrt{1+R}$$, which gives me (-inf, -1) and (1, inf) for branch cuts.

3) Solving $$z+1=-R, z-1=-R$$ gives (-inf, -1) and (-inf, 1), but the overlap cancels(?) to just give (-1,1) as the branch cut.
P: 1,639
 Quote by csnsc14320 OK, so if I try to map just the square roots to the range (-inf, 0) I get:
Ok, lemme' tell you something that is a key to understanding branch-cuts: they're arbitrary. That is, you can extend them from the branch-points any way you like. Now, granted, we try and standardize it somewhat as in the case of sqrt(z) where we just arbitrarly set the cut from -infty to 0 but that's not set in stone. I could equally have it run up the imaginary axis from zero or even a stair-step pattern up to infinity. So in the case of
$$\sqrt{1-z^2}$$

you should look at that and immediately conclude the branch points are at -1 and 1 and the branch-cuts can basically extend anyway from those points, like I said, even stair-steps going up from 1 and stairsteps going down from -1. But to make it simple, we just extend the cut from -1 to 1 (branch-point to branch-point) 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 branch-point.