# Math Help - Branch cut

1. ## Branch cut

Define a branch of the function
$f(z) = ( {z-1 \over z+1})^ {1 \over 3}$

that is regular everywhere except for a cut along the real interval [-1,1] and that takes real positive values for real z>1.

I understand that a branch cut restricts a multi-valued function to a single-valued function, but won't there be several branches here?

2. Probably a better way to describe this, but I think this is a good start:

Some multifunctions have analytic sheets outside a branch cut, others do not. The function $\sqrt[3]{\frac{z-1}{z+1}}$ does and we can show this by considering an arbitrary point $z$ off the branch cut and writing it as:

$1+r_1 e^{i\theta_1}=z,\quad -\pi\leq \theta_1<\pi$

$-1+r_2 e^{i\theta_2}=z,\quad 0\leq \theta_2<2\pi$

That gives the branch-cut between the singular points. Now, $z-1=r_1 e^{i\theta_1}$ and $z+1=r_2 e^{i\theta_2}$. When I substitute that into the radical, I get:

$\left(\frac{r_1}{r_2}\right)^{1/3} e^{i/3(\theta_1+\theta_2+2k\pi)},\quad k=0,1,2$

Draw a picture of this and take a contour say r=3 and go completely around the origin and study how the values of $\theta_1, \theta_2$ and $k$ must change in order for the argument $\theta_1-\theta_2+2k\pi$ to vary smoothly around the circle: Since you want the branch that is real for real z, then we start with $k=0$ at the point z=3. Now go around to the point $z=-3$. The argument in the second quad is then $\pi-\pi=0$ but into the third quadrant, it would become $-\pi-\pi=-2\pi$, so at that point, we add the $2\pi$ to bring it back to zero to assure continuity. Keep going around back to the point $z=3$. In the fourth quad, the argument is $2\pi+0=2\pi$ but once it move into the first quad, it jumps back to $0+0=0$ so we then remove the $2\pi$ to bring the argument back to zero at $z=3$.
So the branch would be:

$

I'm describing the red sheet of the (total) real surface of this multifunction I plotted in the first plot below. Note how this sheet is contiguous and single-valued outside the branch cut.

Contrast this analysis with the function $\sqrt[3]{(z-1)(z+2)}$. There is no way to vary k so that the argument goes smoothly back to zero at the point $z=3$ in a $2\pi$ trip around the origin. In this case, this function does NOT have analytic sheets outside the branch cut $(-1,1)$ but rather consists of a single sheet wrapped around the origin 3 times as seen by the second plot. Note if you follow a circular path around the second plot, it does not return to the same point after a $2\pi$ trip.

3. I'm not entirely satisfied with my analysis above and feel there must be a more simple way of describing the analytic branches of the function. You did check that thread I posted in the other one right:

S.O.S. Mathematics CyberBoard :: View topic - Branches of (z-1)^a (z+1)^b

Also, the branches can be described by differentiating the expression $f(z)=\sqrt[3]{\frac{z-1}{z+1}}$ with respect to r and t for $z=re^{it}$. This gives for example:

$\frac{df}{dt}=1/3\frac{f(t)}{\frac{z-1}{z+1}}\frac{d}{dt}\left(\frac{z-1}{z+1}\right)$

and a similar one for $\frac{df}{dr}$ and any IVP of those equations outside the branch cut will describe the branch and is in fact how I plotted the surfaces of this multifunction above.

Personally, if this was a final-exam question, I'd submit the two differential equations in terms of IVP's (and Opalg's a-b argument for analyticity in the other thread) as the best description of the branches.

4. Have you seen notation like $-x \pm i0$, to denote the points just above/below the branch cut?

For example, if I had the function f(z) = ln(z+1) - ln(z-1) and so I have
$z-1=r_1 e^{i \theta_1} \quad z+1=r_s e^{i \theta_2}$
z-1 has a branch cut from + 1 to $-\infty$ and z+1 from -1 to $-\infty$ so look at the combined cut from +1 to $-\infty$, and look at whether all the branch cut is needed..how would I do that?