# Branch cuts in complex analysis

1. ### csnsc14320

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).

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3. ### clamtrox

939
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.

4. ### csnsc14320

57
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?

5. ### jackmell

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.

6. ### clamtrox

939
Are you using √(-1) = i or √(-1) = -i?

7. ### csnsc14320

57
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.

8. ### jackmell

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.