+0  
 
-1
58
2
avatar

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

 May 20, 2022
 #1
avatar+26287 
+1

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

 

\(\begin{array}{|lrcll|} \hline & z^2 + z + 1 &=& 0 \\ \Rightarrow & z + 1 &=& -z^2 \\ \Rightarrow & z^2 + z &=& -1 \\ \hline \end{array} \begin{array}{|rcll|} \hline z^2 + z + 1 &=& (1+z)^2-z \quad | \quad z + 1 = -z^2 \\ z^2 + z + 1 &=& (-z^2)^2-z \\ z^2 + z + 1 &=& z^4-z \quad | \quad z^2 + z + 1 = 0 \\ 0 &=& z^4-z \\ \mathbf{z^4} &=& \mathbf{z} \\ \hline \end{array}\)

 

\(\begin{array}{|rcll|} \hline z^{49} &=& z^{48}z \\ z^{49} &=& \left(z^{4}\right)^{12}z \quad | \quad z^4 = z \\ z^{49} &=& \left(z\right)^{12}z \\ z^{49} &=& \left(z^{4}\right)^{3}z \quad | \quad z^4 = z \\ z^{49} &=& \left(z\right)^{3}z \\ z^{49} &=& z^{4} \\ \mathbf{z^{49}} &=& \mathbf{z} \\ \hline \end{array}\)

 

\(\begin{array}{|rcll|} \hline && z^{49} + z^{50} + z^{51} + z^{52} + z^{53} \\ &=& z^{49}\left(1 + z + z^2+ z^3 + z^4\right) \\ && \boxed{ z^{49}=z \\ 1 + z + z^2 = 0 \\z^4 = z } \\ &=& z\left(0+ z^3 + z\right) \\ &=& z^4+z^2 \quad | \quad z^4 = z \\ &=& z+z^2 \quad | \quad z^2 + z = -1 \\ &=& -1 \\ \hline \end{array}\)

 

\(\boxed{z^{49} + z^{50} + z^{51} + z^{52} + z^{53} = -1}\)

 

laugh

 May 20, 2022
 #2
avatar+9457 
0

\(z^2 + z + 1 = 0 \\ (z^2 + z + 1)(z - 1) = 0\\ z^3 - 1 = 0\\ z^3 = 1\)

 

So, if z^2 + z + 1 = 0, then z^3 = 1. (Note that the reverse is not true: z^3 = 1 does not imply z^2 + z + 1 = 0.)

 

\(\quad z^{49} + z^{50} + z^{51} + z^{52} + z^{53}\\ = z^{48}(z + z^2 + z^3 + z^4 + z^5)\\ = (z^3)^{16} (z + z^2 + z^3(1 + z + z^2))\\ = 1^{16} (z + z^2 + z^3 (0))\\ = z + z^2 \\ = (z^2 + z + 1) - 1\\ = 0 - 1\\ = -1\)

 May 20, 2022

9 Online Users

avatar