1
$\begingroup$

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?

$\endgroup$
4

2 Answers 2

5
$\begingroup$

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.

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

$\endgroup$
5
  • $\begingroup$ I am confused, how exactly is the GP related of the question being asked? $\endgroup$
    – Aditya
    Dec 12, 2020 at 15:00
  • $\begingroup$ @Aditya I proved that the sum is equal to $z+z^2+..+z^6$ which is GP. $\endgroup$
    – Adola
    Dec 12, 2020 at 15:00
  • $\begingroup$ I understood it now, but you should elaborate on how you got the first line, ie. $z^{100} =z^2$. $\endgroup$
    – Aditya
    Dec 12, 2020 at 15:03
  • $\begingroup$ @Aditya $100$ is in the form of $7k+2$ so $z^{100}=z^2$. $\endgroup$
    – Adola
    Dec 12, 2020 at 15:03
  • 1
    $\begingroup$ 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 $\endgroup$
    – Aditya
    Dec 12, 2020 at 15:04
2
$\begingroup$

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}

$\endgroup$
11
  • $\begingroup$ It's $-1$ since $1$ is not included. $\endgroup$
    – Adola
    Dec 12, 2020 at 14:56
  • $\begingroup$ I don’t understand your signs. Why is there a 7 after the equivalent sign? $\endgroup$
    – Aditya
    Dec 12, 2020 at 15:00
  • $\begingroup$ It is modulo 7. You know conguence relation? $\endgroup$
    – nonuser
    Dec 12, 2020 at 15:01
  • $\begingroup$ No, I am not aware $\endgroup$
    – Aditya
    Dec 12, 2020 at 15:01
  • $\begingroup$ Then just think of $z^{7k+r} = z^r$. $\endgroup$
    – nonuser
    Dec 12, 2020 at 15:02

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .