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Hint:

You're trying to find $z^{53}+z^{52}+z^{51}+z^{50}+z^{49}$, so factor out $z^{49}$ to get $z^{49}(z^4+z^3+z^2+z+1)$. We know what $z^2+z+1$ is, so substitute it in. You now have$z^{49}(z^4+z^3)$, which can be written as $z^{51}(z^2+z)$. Now, we know that $z^2 + z + 1=0$, so $z^2+z=-1$. Substituting, we get $-z^{51}$.

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-24

If

$z^2 + z + 1 = 0$,

find $z^{49} + z^{50} + z^{51} + z^{52} + z^{53}$.

$\begin{array}{|rcll|} \hline \left(z^2 + z + 1 \right)^2 &=& 1+z^2+z^4+2z+2z^2 +2z^3 \\ \left(z^2 + z + 1 \right)^2 &=& 1+z^2+z^4+2z(z^2 + z + 1) \quad | \quad z^2 + z + 1 = 0 \\ \mathbf{0} &\mathbf{=} & \mathbf{1+z^2+z^4} \quad | \quad 0 = z^2 + z + 1 \\ z^2 + z + 1 & = & 1+z^2+z^4 \\ z & = & z^4 \quad | \quad : z \\ 1 & = & z^3 \\ \mathbf{z^3} &\mathbf{=} & \mathbf{1} \\ \hline \end{array}$

$\begin{array}{|lcll|} \hline z^4 = z^3z= 1\cdot z &=& z \\ z^5 = z^3z^2= 1\cdot z^2 &=& z^2 \\ z^6 = \left(z^3\right)^2= 1^2 &=& 1 \\ \hline \end{array}$

In general:

$\begin{array}{|lcll|} \hline z^{0+3n} = 1 \\ z^{1+3n} = z \\ z^{2+3n} = z^2 \\ \hline \end{array}$

$\begin{array}{|rcll|} \hline z^{49} = z^{1+3\cdot16} &=& z \\ z^{50} = z^{2+3\cdot16} &=& z^2 \\ z^{51} = z^{0+3\cdot17} &=& 1 \\ z^{52} = z^{1+3\cdot17} &=& z \\ z^{53} = z^{2+3\cdot17} &=& z^2 \\ \hline \end{array}$

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

$z^2+z+1 = 0\\ (z-1)(z^2+z+1)=0\\ z^3 = 1, z \neq 1\\ z = -\dfrac{1}{2} + \dfrac{i\sqrt3}{2}\\ z^2 = -\dfrac{1}{2} - \dfrac{i\sqrt3}{2}\\$

$z^{49} + z^{50} + z^{51} + z^{52} + z^{53}\\ =z + z^2 + 1 + z + z^2\\ =1+2(-\dfrac{1}{2} + \dfrac{i\sqrt3}{2} + -\dfrac{1}{2} - \dfrac{i\sqrt3}{2})\\ =1 + 2(-1)\\ = -1$