If z = cisΘ, show that cos(nΘ) = 1/2 (z^n + (1/z^n))

I have done:

z^n = (cisΘ)^n = cos(nΘ) + isin(nΘ)

cos(nΘ) = z^n - isin(nΘ)

Don't know where to go from here. Any help or advice would be greatly appreciated.

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- Sep 3rd 2009, 01:22 PMMr.Reesimple complex numbers proof
If z = cisΘ, show that cos(nΘ) = 1/2 (z^n + (1/z^n))

I have done:

z^n = (cisΘ)^n = cos(nΘ) + isin(nΘ)

cos(nΘ) = z^n - isin(nΘ)

Don't know where to go from here. Any help or advice would be greatly appreciated. - Sep 3rd 2009, 01:27 PMMatt Westwood
Now do $\displaystyle z^{-n}$ and add it to what you've already got.

- Sep 3rd 2009, 01:47 PMMr.Ree
I've got it now =]

Thank you very much