Value of expression $\frac{1}{1+z^{8k}+z^{16k}}$

Question

If $z$ is a complex number such that $z^{23}=1.$ Then $\displaystyle \sum^{22}_{k=1}\frac{1}{1+z^{8k}+z^{16k}}$

What I try : Using

$z^{23}=1\Longrightarrow z^{24}=z\Longrightarrow z^{24k}=z^k$

So $\displaystyle \sum^{22}_{k=1}\frac{z^{8k}-1}{(1+z^{8k}+z^{16k})\cdot (z^{8k}-1)}$

$\displaystyle =\sum^{22}_{k=1}\frac{z^{8k}-1}{z^{24k}-1}$

$\displaystyle =\sum^{22}_{k=1}\frac{z^{8k}-1}{z^k-1}$

$\displaystyle =\sum^{22}_{k=1}\bigg(1+z^{k}+z^{2k}+\cdots +z^{7k}\bigg)$

I did not understand How I solve it because when Directly calculate each term with summation is very Complex

Please have a look on that Problem, Thanks

Check this answer https://math.stackexchange.com/a/971846/42969, the same approach works here. — Martin R, Feb 20 '24 at 08:53
For integer $a \not\equiv 0\pmod {23}$, $\sum\limits_{k=1}^{22} z^{ak} = \frac{z^a - z^{23a}}{1 - z^a} = \frac{z^a - 1}{1-z^a} = -1$ — achille hui, Feb 20 '24 at 09:03
Does this answer your question? What is the value of $\sum_{m=1}^{19} \frac{1}{\zeta^{3m}+\zeta^{2m}+\zeta^{m}+1}$ with $\zeta=e^{2\pi i/19}$? – It is the same problem only with different numbers, and the solution method is the same. — Martin R, Feb 20 '24 at 09:12

Abezhiko · Answer 1 · 2024-02-22T14:14:34.000

2

Going on from where you stopped, one has : $$ \begin{align} \sum^{22}_{k=1} \left(1+z^{k}+z^{2k}+\cdots +z^{7k}\right) &= 22 + \sum_{n=1}^7 \sum^{22}_{k=1} z^{nk} \\ &= 22 + \sum_{n=1}^7 \sum^{22}_{k=1} \left(\frac{z^{23n}-1}{z^n-1}-1\right) \\ &= 22 + \sum_{n=1}^7 (-1) \\ &= 15 \end{align} $$

edited Feb 22 '24 at 14:14

answered Feb 21 '24 at 18:33

Abezhiko

8,153

You're right, my bad; I didn't notice that the case $n = 0$ doesn't represent a geometric sequence at first glance. Now it is corrected. – Abezhiko Feb 22 '24 at 14:15

Value of expression $\frac{1}{1+z^{8k}+z^{16k}}$

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