Show that (z^5 - 1)/(z - 1) = z^4 + z^3 + z^2 + z + 1

Start off with your right hand side and consider the sum of a GP.

Also show that (z^5 - 1)/(z - 1) = (z^2 - 2zcos(2pi/5) + 1)(z^2 - 2zcos(4pi/5) + 1)

On the left hand side, we have (z^5 - 1)/(z-1). Try to factorise (z^5 - 1) into the form (z - a)(z - b)(z - c)(z - d)(z - e) where a, b, c, d and e are the zeros of (z^5 - 1) (note: to obtain the zeros of (z^5 - 1), we just set (z^5 - 1) to 0, in other words, z^5 - 1 = 0, which is z^5 = 1, which is root of unity!). From here, you should be able to apply the identity , and hence, obtain your right hand side. I hope this helps

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