Branch cut

Robert Israel — Wed, 22 Jun 2011 04:40:14 Z

To obtain a square root function where the branch cut is at arg(z) = t, you can try

> mysqrt:= (t,z) -> sqrt(-z*exp(-I*t))/sqrt(-exp(-I*t));

sqrt(z^4 - E)

Robert Israel — Wed, 22 Jun 2011 09:57:22 Z

You only have residues at isolated singularities, not at branch points.

Let E = b^4 with b > 0.
In this case, I don't see how "any branch cut that is not on the real axis would be fine", but if you take your function as z* sqrt(b^2+z^2) * sqrt(1 - b^2/z^2) , with the principal branch of sqrt, you'll have branch cuts on the line segment [-b,b] of the real axis and [b,infinity)*I and (-infinity, -b]*I on the imaginary axis. You could then take a contour such as the ellipse (x/a)^2 + (y/c)^2 = 1 with a > b > c and not meet a branch cut.

Completing the calculation

Alec Mihailovs — Wed, 22 Jun 2011 22:56:20 Z

Substitution z=b*w reduces that to the case with b=1. Now, taking the contour close to the real axis, (x+I*y), sqrt((x^2+I*y)^2+1), and sqrt(1-1/(x+I*y)^2) with small y have the same sign of the imaginary part as the sign of y, and the first 2 of these imaginary parts are small, so that product also will have the same sign of the imaginary part as the sign of y.

The sign of the imaginary part of sqrt(z^4-1) is positive for real z on the interval between -1 and 1. That means that the integral over the bottom part of the contour (close to the real axis), oriented counterclockwise, will be close to the -int(sqrt(x^4-1),x=-1..1), and the same for the integral over the top part of that contour. Taking the limit, our contour integral (with b=1) equals

-2*int(sqrt(x^4-1),x=-1..1)=-4*int(sqrt(x^4-1),x=0..1);

                                  1/2
                   1/2           2
           -4/3 I 2    EllipticK(----) = -I Beta(1/4, 3/2)
                                  2

which we can also express in terms of Γ,

convert(rhs(%),GAMMA);

                                   3/2  1/2
                          -2/3 I Pi    2
                          -----------------
                                       2
                             GAMMA(3/4)

And in the general case the answer will be that expression multiplied by b^3 = E^(3/4).

If the originally chosen function had the opposite sign, i.e. if it was -z*sqrt(b^2+z^2)*sqrt(1-z^2/b^2), the sign of the (imaginary part of the) contour integral would be opposite as well.

Alec

PS By the way, this expression of EllipticK(1/sqrt(2)) through GAMMA(3/4) seems to be unknown to simplify,

simplify(2*EllipticK(sqrt(2)/2)*GAMMA(3/4)^2/Pi^(3/2));

                                 1/2
                                2               2
                    2 EllipticK(----) GAMMA(3/4)
                                 2
                    -----------------------------
                                  3/2
                                Pi

evalf(%);

                             0.9999999994

The conversion to hypergeom works though,

simplify(convert(EllipticK(1/sqrt(2)),hypergeom));

                                    3/2
                                  Pi
                           1/2 -----------
                                         2
                               GAMMA(3/4)

but simplify doesn't use this conversion itself, even if the option hypergeom is added to it.

I'm not actually suggesting to add that to simplify - but, perhaps, a new FullSimplify command could be added, working longer, perhaps, than simplify, but trying various combinations of possible simplifications.

Alec

MaplePrimes - answers and comments on Question, How do I change the branch cut definition for the sqrt function?

Branch cut

sqrt(z^4 - E)

Completing the calculation