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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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7 Answers 7

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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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  • $\begingroup$ I see... Thank you for the quick answer! $\endgroup$ Jan 1, 2012 at 11:54
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    $\begingroup$ $\displaystyle{z^{2^k}+\frac{1}{z^{2^k}}=-1}$ for all $k$. $\endgroup$ Jan 1, 2012 at 11:54
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    $\begingroup$ @Jonas: nice observation! $\endgroup$
    – Paul
    Jan 1, 2012 at 12:06
  • $\begingroup$ Dickson polynomials FTW! $\endgroup$ Jan 1, 2012 at 14:20
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    $\begingroup$ Another way to justify that $z+\dfrac1{z}=-1$ is that due to the coefficient symmetry, if $z$ is a root, then $\dfrac1{z}$ is a root as well. By Vieta, then, the sum of those two roots ought to be the negative of the coefficient of the linear term... $\endgroup$ Jan 1, 2012 at 15:15
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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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    $\begingroup$ Very clever! :D $\endgroup$ Jan 1, 2012 at 12:08
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    $\begingroup$ Very nice! If you had not posted this already, I would have used the sum of geometric series (instead of cyclotomic polynomials) to write $$0=z^2+z+1=\frac{z^3-1}{z-1}\Rightarrow z^3=1$$ and $$z^4+\frac{1}{z^4}=(z^3)z+\frac{1}{(z^3)z}=z+\frac{1}{z}=\frac{z^2+1}{z}=\frac{-z}{z}=-1.$$ $\endgroup$ Jan 1, 2012 at 15:48
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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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    $\begingroup$ What would be wrong with systematic reduction using the minimal polynomial? Nothing! $\endgroup$ Jan 1, 2012 at 12:25
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Here is an alternative approach: let's 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} $ we know $\rm\ yz\ (=\: 1)\:$ & $\rm\ y+z\ (=\: z^{-1}\!+z\: =\: -1)\ $

Thus we know $\rm\ \ \ y^2 + z^2\ =\ (y\ +\ z)^2-\ 2\:(y\:z)$

hence we 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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    $\begingroup$ To downvoter: if something is not clear then please feel welcome to ask questions and I will be happy to elaborate. $\endgroup$ Jan 1, 2012 at 22:52
  • $\begingroup$ More generally see here $\endgroup$ Jun 16, 2022 at 7:43
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$$z^2+z+1=0$$ Clearly $z \not= 0$ and therefore $z+\dfrac{1}{z}=-1.$ Note that $$z^{n+1}+\frac{1}{z^{n+1}}=\left(z+\frac{1}{z}\right)\left(z^n+\frac{1}{z^n}\right)-\left(z^{n-1}+\frac{1}{z^{n-1}}\right).$$ Therefore if we define the function $U:\Bbb{N}\cup\{0\} \to \Bbb{C}$ as $$U_n=z^n+\frac{1}{z^n}$$ then we have $U(0)=2,$ $U(1)=-1$ and $$U(n+1)+U(n)+U(n-1)=0\,\,\,\,\,\,\,\,\,\forall n \in \Bbb{N}.$$
Using this recurrence you can calculate $z^n+\dfrac{1}{z^n}$ for any $n \in \Bbb{N}.$

Alos you can follow this approach.

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