Hi, I'm looking at a question from my Pure 6 textbook (united kingdom), it's not actually for homework but I'd like to figure it out. First part of the question goes like this: If 2 cos θ = z + z^-1 prove that (if n is a positive integer) 2 cos n θ = z^n + z^-n. I can get a solution as follows: Let z = cos θ + i sin θ then z^-1 = cos (-θ) + i sin (-θ) by de Moivre's theorem so z^-1 = cos θ - i sin θ so z + z^-1 = 2 cos θ similarly z^n = cos nθ + i sin nθ z^-n = cos -nθ + i sin -nθ z^-n = cos nθ - i sin nθ so z^n + z^-n = 2 cos n θ However I'm not sure if this is valid because of the first line where I let z=cos θ + i sin θ. The question does not state what z is, so can it be assumed to be any complex number? If so then I don't think z=cos θ + i sin θ is valid because that only works when the modulus of z is 1, right? Or can cos θ + i sin θ represent complex numbers of modulus other than 1, maybe if θ itself is complex? But then would de Moivre's theorem still be appliciable? Or does the fact that the question states 2 cos θ = z + z^-1 imply that z must be representable by cos θ + i sin θ ? Thanks very much for any help with this.