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

 Nov 21, 2020
 #1
avatar+12445 
+2

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

 

Hello Guest!

 

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

laugh  !

 Nov 21, 2020
 #2
avatar+9333 
+4

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 

 Nov 21, 2020
edited by hectictar  Nov 21, 2020
 #3
avatar+26213 
+7

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)

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

 

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

 

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

 

laugh

 Nov 21, 2020
 #4
avatar+114914 
+4

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

 Nov 21, 2020
 #5
avatar+114914 
+1

Sorry,

I started mine hours ago and got called away.

I did not know that Heureka and Hectictar had already answered it.

Melody  Nov 21, 2020
 #6
avatar+121047 
+1

Nice work hectictar, heureka and Melody....

 

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

 

 

cool cool cool

 Nov 21, 2020

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