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Given that $z^2 + z + 1 = 0$ where $z$ is a complex number, how do I proceed in calculating $z^4 + \dfrac1{z^4}$? Calculating the complex roots and then the result could be an answer I suppose, but it's not quite elegant. What alternatives are there? |
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From $z^2 + z + 1 = 0$, we have $$z +\frac{1}{z}=-1.$$ Taking square, we get $$z^2 +\frac{1}{z^2}+2=1,$$ which implies that $$z^2 +\frac{1}{z^2}=-1.$$ I think you will know what to do next. |
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Essentially the same calculation also follows from the observation that $x^2+x+1=\phi_3(x)$ is the third cyclotomic polynomial. So $z^3-1=(z-1)(z^2+z+1)=0$, and hence $z^3=1$ for any solution $z$. Therefore $$ z^4+\frac{1}{z^4}=z\cdot z^3+\frac{(z^3)^2}{z^4}=z+z^2=-1. $$ |
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$$\begin{align*} z^4+\frac1{z^4}&=(-z-1)^2+\frac1{(-z-1)^2}\\ &=z^2+2z+1+\frac1{z^2+2z+1}\\ &=(-z-1)+2z+1+\frac1{(-z-1)+2z+1}\\ &=z+\frac1{z}=\frac{z^2+1}{z}=\frac{-z-1+1}{z}=-1 \end{align*}$$ |
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Here is an alternative approach: Lets consider $z^{8}+1$ , and then divide by $z^{4}.$ By using geometric series notice that $$z^{8}+z^{7}+z^{6}+z^{5}+z^{4}+z^{3}+z^{2}+z+1=\left(z^{2}+z+1\right)\left(z^{6}+z^{3}+1\right)=0.$$ Now as $z^{2}+z+1=0$,we know that both $z^{7}+z^{6}+z^{5}=0$ and $z^{3}+z^{2}+z=0$ , and hence $$z^{8}+z^{4}+1=0$$ so that $z^{8}+1=-z^{4}.$ Thus we conclude that $$z^{4}+\frac{1}{z^{4}}=-1.$$ |
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Different people see different things in the relation $z^2+z+1=0$. Just as Jyrki did, I see the third cyclotomic polynomial. Its roots are $-1/2 \pm \sqrt{-3}/2$, the two primitive cube roots of unity. Call one of these $\omega$ and see that $\omega^3=1$, so that $\omega^4=\omega$ and $\omega^{-4}=\omega^2$. Their sum is $-1$, from the defining relation. |
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Hint $\ $ Exploit innate symmetry: for $\rm\ y = z^{-1} $ you know $\rm\ yz\ (=\: 1)\:$ and $\rm\ y+z\ (=\: z^{-1}\!+z\: =\: -1)\ $ Thus you know $\rm\ \ \ y^2 + z^2\ =\ (y\ +\ z)^2-\ 2\:(y\:z)$ hence you know $\rm\ y^4 + z^4\ =\ (y^2\! + z^2)^2 - 2\:(y\:z)^2$ For more on symmetric polynomials see the Wikipedia article on Newton's identities. |
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