📄 dccos.f
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function dccos(z1) double complex dccos, z1, z2 double precision dummy(2), a, b, c, d equivalence (a,z2,dummy(1)), (b,dummy(2)) z2 = z1 c = dexp(b) d = 1.0d0/c b = dsin(a)*(d-c)/2.0d0 a = dcos(a)*(c+d)/2.0d0 dccos = z2 return end
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