If n is a natural number then find the value of $ z^{2012n} + z^{1006n} + 1/z^{2012n} +1/z^{1006n} $ is equal to.
I tried rewriting it as $ t^2+t-1=0 $ where $ t=z+1/z $ and then find roots but I don't know how to use it to get required value.
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3$z^4+z^3+z^2+z+1=0$ so what is $z$? – Thomas Andrews Nov 20 '15 at 18:32
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Hint. Rewrite the condition on $z$ as
$$ \frac{z^4+z^3+z^2+z+1}{z^2} = \frac{1-z^5}{z^2(1-z)} = 0 $$
What does that tell you about $z$? Now consider the possible values of $2012n, 1006n, -2012n, -1006n$ modulo $5$.
Brian Tung
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HINT:
As $z\ne0,$ multiply throughout by $z^2$
$$z^4+z^3+z^2+z+1=0\implies z^5-1=(z-1)(z^4+z^3+z^2+z+1)=0\implies z=e^{2\pi m i/5}$$ where $m\equiv1,2,3,4\pmod5$
lab bhattacharjee
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Can I take n=1 and then divide all powers by 5 and write remainder, which will give $ z^2 +z + 1/z^2 + 1/z $ ? – Mathematics Nov 21 '15 at 15:39