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Math Help - Branch cut

  1. #1
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    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?
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  2. #2
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    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:

    <br />
\sqrt[3]{\frac{z-1}{z+1}}=\begin{cases}\left(\frac{r_1}{r_2}\right)  ^{1/3} e^{i/3(\theta_1-\theta_2)},\quad \text{Im}(z)>0 \\ \left(\frac{r_1}{r_2}\right)^{1/3} e^{i/3(\theta_1-\theta_2+2\pi)},\quad \text{Im}(z)<0\end{cases}

    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.
    Attached Thumbnails Attached Thumbnails Branch cut-cube-real-sheet-1.jpg   Branch cut-cube-imag-sheet-2.jpg  
    Last edited by shawsend; October 16th 2009 at 04:25 PM. Reason: added second plot.
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  3. #3
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    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.
    Last edited by shawsend; October 18th 2009 at 06:12 AM.
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  4. #4
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    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?
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