<s>[INST] Complex number $z=cis(\frac{2k\pi}{5})$ for any integer $k$ such that $z \ne 1$? Consider the complex number $z=cis(\frac{2k\pi}{5})$ for any integer $k$ such that $z \ne 1$.
(a) Show that $z^n+\frac{1}{z^n}=2\cos(\frac{2nk\pi}{5})$ for any integer $n$.
(b) Show that $z^5=1$. Hence, or otherwise, show that $1+z+z^2+z^3+z^4=0$.
(c) Find the value of $b$, given that $(z+\frac{1}{z})^2+(z^2+\frac{1}{z^2})^2=b$.
My approach:
(a) This was easy, I rewrote this as $z^n+z^{-n}$, applied De Moivre's theorem, the sines cancelled out and that was all.
(b) I proved the first part that $z^5=1$, but have trouble with the second part ("hence, or otherwise...") - I tried rewritting a term, say the term $z$, as $cis(\frac{2k\pi}{5}))=cis(\frac{1}{5}*2k\pi)=(cis(2k\pi))^{1/5}=1^{1/5}=1$. This, I thought, will make all the terms be equal to 1. But the first sentence of the exercise says "for any integer $k$ such that $z\ne1$" - but how is it even possible? No matter what value of $k$ I substitute, it will always be a multiple of $2\pi$ so the $cis$ will be 1. What am I missing? How do I solve that?
(c) I think that I should use (a) here but I got stuck.
 [/INST] Observe that
$$z^5=1\iff 0=z^5-1=(z-1)(z^4+z^3+z^2+z+1)$$
and since $\;z\neq1\;$ you get (b)
For (c) you can indeed use (b), and remember that all the roots of order $\;5\;$ are a cyclic (multiplicative) group, so
$$1+z+z^2+z^3+z^4=1+\overbrace{z}^{=z^{-4}}+z^2+z^{-2}+\overbrace{z^4}^{=z^{-1}}=\left(z+\frac1z\right)^2-1+\left(z^2+\frac1{z^2}\right)^2-2$$
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