Let z = cos u = (e^iu + e^-iu)/2

so that iu = cos ^-1(z)

(e^iu)^2 - 2z(e^iu) + 1 = 0

positive solution

e^iu = z + (z^2 - 1)^(1/2)

i*ln(e^u) = iu = ln[z + (z^2 - 1)^(1/2)]

u = cos^-1 (z) = -i[log(z+(z^2-1)^1/2)]?

Regards - Ian

OR see a really cool maths proof at

http://i869.photobucket.com/albums/ab253/iansurnam...