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$z$ is a complex number and $z^2+z+1=0$.

$$z^{10}+\frac{1}{z^{10}}=?$$

For the solution:

  1. the roots of $z^2+z+1$ are: $z_1=-\frac12+\frac{\sqrt3}{2}i$ and $z_2=-\frac12-\frac{\sqrt3}{2}i$
  2. converting these to their trigonometrical forms, we get: $z_1=\cos\frac{2\pi}{3}+i\sin\frac{2\pi}{3}$ and $z_2=\cos\frac{7\pi}{6}+i\sin\frac{7\pi}{6}$
  3. How do I proceed?
Arctic Char
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user176791
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5 Answers5

15

Hint: $z^2+z+1=0$ implies that $0=(z-1)(z^2+z+1)=z^3-1$, so $z^3=1$. Also $z^{10}=z^9\cdot z$ and $z^{-10}=z^2\cdot z^{-12}$.

Jyrki Lahtonen
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  • Nice observation :D (+1) – AlexR Sep 30 '14 at 20:00
  • That's quite the trick. – qwr Sep 30 '14 at 20:11
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    If you don't notice the trick, you can proceed this way: Given that $z^2+z+1=0$, $z$ cannot be zero. Divide by $z$ to get $(z+1+z^{-1})=0$, from which $z+z^{-1}=-1$. Then $(z+z^{-1})^2=z+2+z^{-2}=1$, so $z^2+z^{-2}=-1$. Now $1=(-1)(-1)= (z+z^{-1})(z^2+z^{-2})=z^3+z+z^{-1}+z^{-3}=z^3-1+z^ {-3}$, so $z^ {-3}+z^3=2$. You can get to the answer from here. – Steve Kass Sep 30 '14 at 21:28
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Using $e^{i x} = \cos(x) + i \sin(x)$ then for $z_{1} = e^{2 \pi i/3}$ it is seen that \begin{align} z_{1}^{10} + \frac{1}{z_{1}^{10}} &= e^{20 \pi i/3} + e^{- 20 \pi i/3} = e^{6 \pi i + 2 \pi i/3} + e^{-6 \pi i - 2\pi i/3} \\ &= e^{2\pi i/3} + e^{-2\pi i/3} = 2 \cos(2\pi /3) = -1 \end{align}

Leucippus
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    There is an error with $z_2$ in the OP, thus your result for $z_2$ is erroneous since $z_2^2 + z_2 + 1 \ne 0$. – AlexR Sep 30 '14 at 20:10
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Hints:

  • Note $z_1z_2=1$

  • You could write your (2) as $z_1=e^{i2\pi/3}$ and $z_2=e^{-i2\pi/3}$.

  • Taking the $10$th power of an exponential is easy

Henry
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The solutions to $z^2 + z + 1 = 0$ are the complex $z = 1 \angle (\pm 1/3)$.

So $$\begin{align} z^{10} + z^{-10} & = \left(1\angle (\pm 1/3)\right)^{10} + \left(1\angle (\pm 1/3)\right)^{-10} \\& = \left(1\angle (\pm 10/3)\right) + \left(1\angle (\mp 10/3)\right) \ \\ & = (1\angle 10/3) + (1\angle -10/3) \\ & = (1\angle 1/3) + (1\angle -1/3) \\ & = -1 \end{align}$$

DanielV
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  • What does $;\angle;$ mean here? – Timbuc Oct 01 '14 at 02:33
  • It's old fashioned polar notation. I use it because I hate the verbose Euler form. The angle is in revolutions. – DanielV Oct 01 '14 at 04:15
  • I still don't get that notation. The roots of $;z^2+z+1=0;$ are $$\frac{-1\pm\sqrt{-3}}2=e^{\pm2\pi i/3}=\text{cis},\left(\frac{2\pi i}3\right)$$ the last notation being the one used here in high school. How does your notation means this? – Timbuc Oct 01 '14 at 09:27
  • Convert Radians to revolutions. $2\pi/3 \text{ radians} = 1/3 \text{ rev}$. Then it is $\text{Magnitude } \angle \text{ revolutions}$. It's an uncommon notation that I prefer, but I didn't actually expect any response, I really expected this response to be ignored. Aside, $$\frac{-1 \pm \sqrt{-3}}{2} = e^{\pm 2\pi i /3}$$ is not a notation, and it is a definition that depends strongly on the use of radians. – DanielV Oct 01 '14 at 15:05
  • Well, yes: radians are the usual stuff, by far, used in mathematis and physics (and most other sciences I am aware of). – Timbuc Oct 01 '14 at 15:14
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$z^2+z+1 = 0 \implies z^2+z = -1 $

$z^2= -z-1 \implies z^3 = -z^2-z = -(z^2+z)= 1$

$z^{10} = z^3\cdot z^3\cdot z^3\cdot z= z \implies$

$z^{10} + z^{-10} = z + 1/z = \frac{z^2 +1 }z$ but $z^2+1 = -z$ so this results in $-z/z = -1$

Siong Thye Goh
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axouxere
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