$z$ is a complex number such that $z^7=1$, where $z\not =1$. Find the value of $z^{100}+z^{-100} + z^{300}+z^{-300} + z^{500}+z^{-500}$

Question

Let $z=e^{i\frac{2\pi}{7}}$

Then the expression, after simplification turns to $$2[\cos \frac{200\pi}{7} +\cos \frac{600 \pi}{7} +\cos \frac{1000\pi}{7}]$$

How do I solve from here?

Answer 1

Since $z^7=1$ we have that it's equal to

$$z^2+z^5+z^6+z+z^3+z^4$$

$$=z^6+z^5+z^4+z^3+z^2+z$$ because $z^{7k+1}=z, z^{7k+2}=z^2,...$. This is equal to $-1$ since $z^7-1=(z-1)(z^6+..+1)=0$ and thus $z^6+..+1=0$ noting that $z -1\neq 0$.

Also we can solve using sum of geometric progression.

$\frac{a(r^n-1)}{r-1}=\frac{z(z^6-1)}{z-1}=\frac{z^7-z}{z-1}=\frac{1-z}{z-1}=-1$

Answer 2

Since $$100\equiv _7 2$$ and $$300\equiv _7 -1$$ and $$500\equiv _7 3$$ We have \begin{align} &=z^{2}+z^{-2} + z^{-1}+z^{1} + z^{3}+z^{-3} \\ &= {z^5+z+z^2+z^4+z^6+1\over z^3}\\ & ={z^6+z^5+z^4+\color{red}{z^3}+z^2+z+1-\color{red}{z^3}\over z^3}\\ &= {{z^7-1\over z-1} -z^3\over z^3} \\ &= {0-z^3\over z^3} =-1\end{align}