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$$z_1 \text{ and } z_2 \text{ are the solutions of } 1-z+z^2=0$$
$$E=(z_1^4-z_1^3+2z_1^2-2z_1+1)^{2005}+(z_2^4-z_2^3+2z_2^2-2z_2+1)^{2005}$$

Which is the value of $E$ ?

I have solved the equation:

\begin{align*}\Delta = 1-4=-3=3i^2&\Rightarrow z_{1,2}=\frac{1\pm i\sqrt{3}}{2} \end{align*}

One solution would be to write these numbers in trigonometric form. But I am sure there is an easier way if I write E differently, but I can't find it.

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Start by using long polynomial division with remainder to remove multiples of $z^2-z+1$ from the inner polynomials. –  Henning Makholm Oct 27 '11 at 16:19
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1 Answer

up vote 10 down vote accepted

If you know that $z_i^2-z_i+1 = 0$, then since $$x^4 - x^3 + 2x^2 - 2x + 1 = (x^2-x+1)(x^2+1) - x,$$ it follows that: $$z_1^4 - z_1^3 + 2z_1^2 - 2z_1 + 1 = (z_1^2 - z_1 + 1)(z_1^2 + 1) - z_1 = 0(z_1^2+1) - z_1 = -z_1.$$

Much simpler than trying to work directly with the roots.

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Thank you! $E=-z_1-z_2=-S=-(-\frac{b}{a})=\frac{-1}{1}=-1$ –  Daniel Oct 27 '11 at 16:45
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