Difficult Precalc Problem

If $z^2+z+1=0$,  find  $z^{49} + z^{50} + z^{51} + z^{52} + z^{53}.$

Hello Guest!

$z^2+z+1=0\\ z=-\frac{1}{2}\pm\sqrt{\frac{1}{4}-1}\\ z=-\frac{1}{2}\pm\sqrt{-\frac{3}{4}}\\ z=-\frac{1}{2}\pm \frac{1}{2}\sqrt{3}\cdot i\\ \color{blue}z=-\frac{1}{2}\pm sin(\frac{\pi}{3})\cdot i$

  !

IDK if this is the best way...but we can first simplify it like this:

z⁴⁹ + z⁵⁰ + z⁵¹ + z⁵² + z⁵³

                                                        Factor  z⁴⁹  out of the first three terms and  z⁵¹ out of the last two terms

=  z⁴⁹(1 + z + z²)  +  z⁵¹(z + z²)

                                                        Since  z² + z + 1 = 0,   1 + z + z² =  0   and   z + z²  =  -1

=  z⁴⁹( 0 )  +  z⁵¹( -1 )

=  -z⁵¹

Now by the quadratic formula,  $z\ =\ \frac{-1\pm\sqrt{1^2-4(1)(1)}}{2(1)}\ =\ \frac{-1\pm\sqrt{-3}}{2}\ =\ -\frac12\pm \frac{\sqrt3}{2}i$

Let's pick  $z\ =\ -\frac12 + \frac{\sqrt3}{2}i$     (If we picked  $z\ =\ -\frac12 - \frac{\sqrt3}{2}i$   we would get the same answer)

Now let's re-express  z  to be in the form  $r(\cos\theta+i\sin\theta)$  so that we can use DeMoivre's Theorem.

By the Pythagorean Theorem,

$r^2\ =\ (-\frac12)^2+(\frac{\sqrt3}{2})^2\ =\ 1$   so taking the positive sqrt, we get     $r \ =\ 1$

An angle which has a cos of  $-\frac12$  and a sin of  $\frac{\sqrt{3}}{2}$  is  $\frac{2\pi}{3}$,  so let  $\theta\ =\ \frac{2\pi}{3}$

And so...

$z\ =\ 1(\cos(\frac{2\pi}{3})+i\sin(\frac{2\pi}{3}))$          (we can check in a calculator that this does equal $-\frac12 + \frac{\sqrt3}{2}i$ )

Then by DeMoivre's Theorem,

$z^{51}\ =\ (1)^{51}(\cos(51\cdot\frac{2\pi}{3})+i\sin(51\cdot\frac{2\pi}{3}))\\~\\ z^{51}\ =\ (1)^{51}((1)+i(0))\\~\\ z^{51}\ =\ (1)^{51}\\~\\ z^{51}\ =\ 1$

And so...

$-z^{51}\ =\ -1$

Check 

If $z^2 + z + 1 = 0$, find
$z^{49} + z^{50} + z^{51} + z^{52} + z^{53}$.

1)

$\begin{array}{|rcll|} \hline && \mathbf{z^{49} + z^{50} + z^{51} + z^{52} + z^{53}} \\ &=& \left(z^{49} + z^{50} + z^{51} \right) + z^{52} + z^{53} \\ &=& z^{49}\left(1+ z + z^2 \right) + z^{52} + z^{53} \quad | \quad \mathbf{z^2 + z + 1 = 0} \\ &=& z^{49}\cdot 0 + z^{52} + z^{53} \\ &=& \mathbf{z^{52} + z^{53}} \\ \hline \end{array}$

2)

$\begin{array}{|rcll|} \hline && \mathbf{z^{49} + z^{50} + z^{51} + z^{52} + z^{53}} \\ &=& z^{49} + \left(z^{50} + z^{51} + z^{52}\right) + z^{53} \\ &=& z^{49}+ z^{50}\left(1+ z + z^2 \right) + z^{53} \quad | \quad \mathbf{z^2 + z + 1 = 0} \\ &=& z^{49}+ z^{50}\cdot 0 + z^{53} \\ &=& \mathbf{z^{49} + z^{53}} \\ \hline \end{array}$

$\begin{array}{|rcll|} \hline z^{52} + z^{53} &=& z^{49} + z^{53} \\ z^{52} &=& z^{49} \quad | \quad :z^{49}\\ \dfrac{z^{52}}{z^{49}} &=& 1 \\ z^{52-49} &=& 1 \\ \mathbf{z^{3}} &=& \mathbf{1} \\ \hline \end{array}$

$\begin{array}{|rcll|} \hline \mathbf{z^{49} + z^{50} + z^{51} + z^{52} + z^{53}} &=& \mathbf{z^{49} + z^{53}} \\ &=& z^{3\cdot16+1} + z^{3\cdot17+2} \\ &=& z^{3\cdot16}z + z^{3\cdot17}z^2 \\ &=& \left(z^3\right)^{16}z + \left(z^3\right)^{17}z^2 \quad | \quad \mathbf{z^{3} = 1 } \\ &=& 1^{16}z + 1^{17}z^2 \\ &=& z + z^2 \\ && \boxed{z^2 + z + 1 = 0 \\ z^2 + z = -1 } \\ \mathbf{z^{49} + z^{50} + z^{51} + z^{52} + z^{53}} &=& \mathbf{-1} \\ \hline \end{array}$

Lets see

$z^2 + z + 1 = 0\\~\\ z=\frac{-1\pm \sqrt{1-4}}{2}\\ z=\frac{-1\pm \sqrt{-3}}{2}\\ z^2=\frac{1+-3\mp2\sqrt{-3}}{4}\\ z^2=\frac{-2\mp2\sqrt{-3}}{4}\\ z^2=\frac{-1\mp\sqrt{-3}}{2}\\ z^2=\bar z$

$z^3=z^2*z=\bar z*z=\frac{1}{4}-\frac{-3}{4}=1\\ so\\ z^{3n}=1\qquad \text{where n is a positive integer} so\\ z^{48}=z^{51}=1$

$z^{49}=z\\ z^{50}=\bar z\\ z^{51}=1\\ z^{52}=z\\ z^{53}=\bar z\\~\\ \text{Add them together and get}\\ 1+2(z+\bar z)\\ =1+2(\frac{-1}{2}+\frac{-1}{2})\\ =1+2(-1)\\ =1-2\\ =-1$

LaTex:

z^2 + z + 1 = 0\\~\\
z=\frac{-1\pm \sqrt{1-4}}{2}\\
z=\frac{-1\pm \sqrt{-3}}{2}\\
z^2=\frac{1+-3\mp2\sqrt{-3}}{4}\\
z^2=\frac{-2\mp2\sqrt{-3}}{4}\\
z^2=\frac{-1\mp\sqrt{-3}}{2}\\
z^2=\bar z

z^3=z^2*z=\bar z*z=\frac{1}{4}-\frac{-3}{4}=1\\
so\\
z^{3n}=1\qquad \text{where n is a positive integer}
so\\
z^{48}=z^{51}=1

z^{49}=z\\
z^{50}=\bar z\\
z^{51}=1\\
z^{52}=z\\
z^{53}=\bar z\\~\\
\text{Add them together and get}\\
1+2(z+\bar z)\\
=1+2(\frac{-1}{2}+\frac{-1}{2})\\
=1+2(-1)\\
=1-2\\
=-1

Nice work hectictar, heureka and Melody....

I don't know much about these.....maybe I'll learn something  by  looking at each  solution  !!!!