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Let $\omega_n=e^{2\pi i/n}$. Prove $1+\omega_n+\omega^2_n+\cdots+\omega^{n-1}_n=0.$

Note: I tried using mathematical induction and some basic trig identities but I can't seem to crack this one. Any hints???

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    $\begingroup$ think of it as a geometric sum $\endgroup$ – fGDu94 Feb 10 '20 at 20:59
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    $\begingroup$ Sum of a geometric series? $\endgroup$ – saulspatz Feb 10 '20 at 20:59
  • $\begingroup$ use the fact that the points $\omega_n^k$ for $k=0,1,2,,n-1$ are vertex of a regular polygon of $N$ sides. $\endgroup$ – Emilio Novati Feb 10 '20 at 21:10
  • $\begingroup$ Please try to make the titles of your questions more informative. For example, Why does $a<b$ imply $a+c<b+c$? is much more useful for other users than A question about inequality. From How can I ask a good question?: Make your title as descriptive as possible. In many cases one can actually phrase the title as the question, at least in such a way so as to be comprehensible to an expert reader. You can find more tips for choosing a good title here. $\endgroup$ – Shaun Feb 10 '20 at 22:48
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$$1+\omega_n+\omega_n^2+\cdots+\omega_n^{n-1} = \frac{1-\omega_n^n}{1-\omega_n}=\frac{1-e^{2\pi i}}{1-\omega_n}=\frac{1-1}{1-\omega_n}=0.$$

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  • $\begingroup$ Took me a while, but I did it and it was this way! $\endgroup$ – whitenoise Feb 13 '20 at 22:13
  • $\begingroup$ nth partial sum of a geometric series on LHS of course. $\endgroup$ – whitenoise Feb 13 '20 at 22:18
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    $\begingroup$ Glad you were able to work it out! $\endgroup$ – mjw Feb 14 '20 at 0:41
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HINT: Notice that $\omega_n^n=1$ so $\omega_n$ is a root of $x^n-1$. Also, $$x^n-1=(x-1)(1+x+x^2+...+x^{n-1})$$ Thus, since $\omega_n-1\ne 0$, you can conclude that...

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Note that

$\omega_n = e^{2\pi i / n} \tag 1$

satisfies

$\omega_n^n - 1 = (e^{2\pi i / n})^n - 1$ $= e^{2 \pi n i / n} - 1 = e^{2 \pi i} - 1 = 1 - 1 = 0; \tag 2$

we also have the well-known identity

$\omega_n^n - 1 = (\omega_n - 1)(1 + \omega_n + \omega_n^2 + \ldots + \omega_n^{n - 1})$ $= (\omega_n - 1) \displaystyle \sum_0^{n - 1} \omega_n^k; \tag 3$

combining (2) and (3) yields

$(\omega_n - 1)(1 + \omega_n + \omega_n^2 + \ldots + \omega_n^{n - 1})$ $= (\omega_n - 1) \displaystyle \sum_0^{n - 1} \omega_n^k = 0, \tag 4$

and since

$\omega_n \ne 1, \tag 5$

we have

$\omega_n - 1 \ne 0; \tag 6$

now dividing (4) through by $\omega_n - 1$ yields

$1 + \omega_n + \omega_n^2 + \ldots + \omega_n^{n - 1} = \displaystyle \sum_0^{n - 1} \omega_n^k = 0, \tag 7$

the desired result.

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  • $\begingroup$ Thanks for the detailed reply! What is the name of the identity in line (3)? $\endgroup$ – whitenoise Feb 10 '20 at 21:33
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    $\begingroup$ @whitenoise: I don't know that it has a name but it pops up everywhere. A more general case is $a^n - b^n = (a - b) \displaystyle \sum_0^{n - 1} a^k b^{n - k - 1}$ $\endgroup$ – Robert Lewis Feb 10 '20 at 21:37

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