If $z$ is any complex number and $z^{23}=1$ then evaluate $\displaystyle \sum^{22}_{r=0}\frac{1}{1+z^r+z^{2r}}$
Try: From $$z^{23}-1=(z-1)(1+z+z^2+\cdots +z^{22})$$
And our sum $$\sum^{22}_{r=0}\frac{1}{1+z^r+z^{2r}}=\frac{1}{3}+\frac{1}{1+z+z^2}+\frac{1}{1+z^2+z^4}+\cdots +\frac{1}{1+z^{22}+z^{44}}$$
Now i did not understand how can i simplify it.
Could some help me. Thanks.
On
Define $f(x)=\frac{1}{1+x+x^2}$. Then we have to calculate $$ \sum_{k=0}^{22}f(z^k)=\frac{1}{3}+\sum_{k=1}^{22}f(z^k). $$ Let's work out the polynomial expression for $f$ using $(a+b)(a-b)=a^2-b^2$. $$ f(s)=\frac{s-1}{s^3-1}=\frac{(s-1)(s^3+1)}{s^6-1}=\frac{(s-1)(s^3+1)(s^6+1)}{s^{12}-1}=\frac{(s-1)(s^3+1)(s^6+1)(s^{12}+1)}{s^{24}-1}. $$ Since $s^{24}=s$ for the $23$d roots of unity, we get $$ f(z^k)=(z^{3k}+1)(z^{6k}+1)(z^{12k}+1)=1+z^{3k}+z^{6k}+z^{9k}+z^{12k}+z^{15k}+z^{18k}+z^{21k}.\tag{1} $$ Now as the OP has noticed $$ s^{23}-1=(s-1)(1+\underbrace{s+s^2+\ldots+s^{22}}_{\Phi(s)}). $$ For the $z^k$, $k=1,2,\ldots,22$, it makes (as all those are primitive roots) $$ 0=\underbrace{(z^k-1)}_{\ne 0}(1+\Phi(z^k))\quad\Leftrightarrow\quad \Phi(z^k)=-1. $$ Finally, summing up (1) we get $$ \sum_{k=1}^{22}f(z^k)=22+\Phi(z^3)+\Phi(z^6)+\Phi(z^9)+\Phi(z^{12})+\Phi(z^{15})+\Phi(z^{18})+\Phi(z^{21})=22-7=15. $$ Plus $\frac13$.
If $z^{23}=1$ and $z\not=1$ (otherwise the sum is trivially equal to $23/3$) then $z^k$ is a primitive $23$-th root of unity for any $k=1,2\dots,22$ (note that $23$ is a prime number). Hence $$\begin{align} \sum^{22}_{r=0}\frac{1}{1+z^r+z^{2r}}&=\frac{1}{3}+\sum_{r=1}^{22}\frac{1}{1+z^r+z^{2r}}\\ &=\frac{1}{3}+\sum_{r=1}^{22}\frac{1}{1+z^{8r}+z^{16r}}\\ &=\frac{1}{3}+\sum_{r=1}^{22}\frac{z^{8r}-1}{(1+z^{8r}+z^{16r})(z^{8r}-1)}\\ &=\frac{1}{3}+\sum_{r=1}^{22}\frac{z^{8r}-1}{z^{24r}-1} =\frac{1}{3}+\sum_{r=1}^{22}\frac{z^{8r}-1}{z^{r}-1}\\ &=\frac{1}{3}+\sum_{r=1}^{22}(1+z^r+z^{2r}+z^{3r}+z^{4r}+z^{5r}+z^{6r}+z^{7r})\\ &=\frac{1}{3}+22+7\sum_{r=1}^{22}z^r. \end{align}$$ Can you take it from here?