Branch cuts in complex analysis

In summary: That is, are you supposed to look at the whole expression and figure out the branch-points and cuts for the entire expression? Or are you supposed to look at the terms, with the sqrt, and determine them for those terms? That is, suppose you have:f(z)=z+\sqrt{1-z^2}+\sqrt{z+1}\sqrt{z-1}then what are your branch-points and cuts? I'm not sure what your original question was asking.In summary, the three expressions given can be reduced to the same expression, but they have different branch cuts due to the presence of the square root function. The branch cuts can be extended arbitrarily from the branch points, but are typically standardized in
  • #1
csnsc14320
57
1

Homework Statement


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)

Homework Equations


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).
 
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  • #2
csnsc14320 said:
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.
 
  • #3
clamtrox said:
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?
 
  • #4
csnsc14320 said:

Homework Statement


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:

[tex]f(z)=z+\sqrt{z^2-1}[/tex]

and I know [itex]G(z)=\sqrt{z^2-1}[/itex] is not a polynomial and that the quantity [itex]z^2-1=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 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.
 
  • #5
csnsc14320 said:
2+isqrt(1-2^2) = 2+isqrt(-3) = 2+sqrt(3)

Are you using √(-1) = i or √(-1) = -i?
 
  • #6
jackmell said:
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) [tex]z^2-1=-R[/tex] where R ranges from (0, inf).
Solving for Z yields [tex]z = \pm \sqrt{1-R}[/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]1-z^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, z-1=-R[/tex] gives (-inf, -1) and (-inf, 1), but the overlap cancels(?) to just give (-1,1) as the branch cut.
 
  • #7
csnsc14320 said:
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
[tex]\sqrt{1-z^2}[/tex]

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.

However, I'm not really sure what your particular question is asking.
 
Last edited:

1. What are branch cuts in complex analysis?

Branch cuts are discontinuities in a complex function that occur when there are multiple possible values for the function at a certain point. They are often represented as line segments or curves in the complex plane.

2. How do branch cuts affect the behavior of complex functions?

Branch cuts can cause a complex function to have multiple branches, each with a different set of values. This can lead to unexpected behavior, such as jumps or discontinuities, when the function is evaluated along the branch cut.

3. Can branch cuts be avoided in complex analysis?

In some cases, it is possible to avoid branch cuts by choosing a different branch of a multivalued function. However, in general, branch cuts are a fundamental aspect of complex functions and cannot be completely avoided.

4. How are branch cuts related to branch points in complex analysis?

Branch cuts are often associated with branch points, which are points in the complex plane where a multivalued function becomes single-valued. A branch cut is typically drawn from a branch point to infinity, separating different branches of the function.

5. What is the significance of branch cuts in the study of complex analysis?

Branch cuts are important in complex analysis because they allow us to understand the behavior of multivalued functions and make sense of their complex behavior. They also play a key role in the study of complex integration and the development of powerful mathematical tools, such as the Cauchy-Riemann equations.

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