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

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?

• You can't assume $z=e^{2\pi i/7}$ like that...\ Dec 12, 2020 at 14:51
• Please don't use \frac in exponents or limits of integrals. It looks bad and confusing, and it rarely appears in professional mathematics typesetting. Dec 12, 2020 at 14:52
• Hint: $z^{100}=z^2$ Dec 12, 2020 at 14:52
• Dec 12, 2020 at 16:01

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.

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

• I am confused, how exactly is the GP related of the question being asked? Dec 12, 2020 at 15:00
• @Aditya I proved that the sum is equal to $z+z^2+..+z^6$ which is GP. Dec 12, 2020 at 15:00
• I understood it now, but you should elaborate on how you got the first line, ie. $z^{100} =z^2$. Dec 12, 2020 at 15:03
• @Aditya $100$ is in the form of $7k+2$ so $z^{100}=z^2$. Dec 12, 2020 at 15:03
• Yes, as I said I understood, I just suggested that you should put it in your answer because it doesn’t immediately click. Thanks for the help Dec 12, 2020 at 15:04

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}

• It's $-1$ since $1$ is not included. Dec 12, 2020 at 14:56
• I don’t understand your signs. Why is there a 7 after the equivalent sign? Dec 12, 2020 at 15:00
• It is modulo 7. You know conguence relation? Dec 12, 2020 at 15:01
• No, I am not aware Dec 12, 2020 at 15:01
• Then just think of $z^{7k+r} = z^r$. Dec 12, 2020 at 15:02