complex(kind(1.d0)) function ztan (z)

        implicit none
  c march 1979 edition.  w. fullerton, c3, los alamos scientific lab.

        complex(kind(1.d0)) z
        real(kind(1.d0)) d1mach,eps, xmax, ylarge, ybig, ymin,sn2x,den
        real(kind(1.d0)) x,y,x2,y2

        integer irold,irold2
        external d1mach

        data eps, xmax, ylarge, ybig, ymin / 4*0.0d0, 1.50d0 /

  c
        if (eps.ne.0.0) go to 10
        eps =  d1mach(4)
        xmax = 1.0/eps
        ylarge = -0.5*log(0.5*d1mach(3))
        ybig = -0.5*log(0.5*sqrt(d1mach(3)))
  c
   10   x = real(z)
        y = aimag(z)
        if (abs(y).gt.ylarge) go to 30
  c
        x2 = 2.0*x
        y2 = 2.0*y
        if (abs(x2).gt.xmax) go to 20
  c
        call entsrc (irold, 1)
        sn2x = sin(x2)
        call erroff
        den = cos(x2) + cosh(y2)
        call erroff
        call entsrc (irold2, irold)
  c
        if (den.lt.x*x2*eps*eps) call seteru ( 72hztan    tan is nearly si
       1ngular, x is near pi/2 or 3*pi/2 and y is near 0, 72, 3, 2)
        if (den.lt.x*x2*eps) call seteru (74hztan    answer lt half precis
       1ion, x is near pi/2 or 3*pi/2 and y is near 0, 74, 1, 1)
  c

        ztan = cmplx (sn2x/den, sinh(y2)/den, kind(1.d0))

        return
  c
   20   if (abs(y).lt.ymin) call seteru (75hztan    answer would have no p
       1recision, abs(x) is very big and abs(y) small, 75, 4, 2)
        if (abs(y).lt.ybig) call seteru (69hztan    answer lt half precisi
       1on, abs(x) is very big and abs(y) small, 69, 2, 1)
  c

        ztan = cmplx (0.0, tanh(y2), kind(1.d0))

        return
  c

   30   ztan = cmplx (0.0, -sign(1.0d0, y), kind(1.d0))

        return
  c
        end