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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.

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# Homework Help: Complex number problem

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