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Consider a sequence $a_n$ $$a,b,c,a,b,c,\ldots$$ and we want to find an expression/formula to represent this.
I'm new to complex numbers, and it feels we can somehow use the expression $\left(e^{i2\pi/3}\right)^n$ to represent above sequence. I've been trying, but I'm not able to form a connection between the two. It seems I need to scale the complex numbers successively by $\dfrac{a}{e^{i0}},\dfrac{b}{e^{i2\pi/3}},\dfrac{c}{e^{-i2\pi/3}} $. I'm kinda stuck.. Appreciate any help...

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Let's just denote $e^{2i\pi/3}$ by $\omega$.

The main observation is that $\omega^n + \omega^{2n} + \omega^{3n}$ is $3$ when $n$ is divisible by $3$ and $0$ otherwise.

Also note that $\omega^3 = 1$.

So you can express your sequence as: $$\frac a3 (\omega^n + \omega^{2n} + 1) + \frac b3 (\omega^{n-1} + \omega^{2n-2} + 1) + \frac c3 (\omega^{n-2} + \omega^{2n-4} + 1)$$

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  • $\begingroup$ I'm sure you mean $\omega^n + \omega^{2n} + \omega^{3n} = 3$ when $n=3k$ ? $\endgroup$ – rsadhvika Jun 20 '18 at 6:15
  • $\begingroup$ Oops, lemme change it $\endgroup$ – Kenny Lau Jun 20 '18 at 6:16
  • $\begingroup$ Oh I know that $1+w+w^2=0$, but it seems more general than this. If $w^n=1$ and when $n\not | ~m$, we have $$\sum\limits_{k=0}^{n-1}(w^m)^{k} = 0$$ I'll google for a proof and try digesting this first... Thanks again :) $\endgroup$ – rsadhvika Jun 20 '18 at 6:41
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The sequence defined by $$z_n=r+s\exp(2\pi ni/3)+t\exp(4\pi ni/3)$$ has period $3$. All you need to do is to fins $r$, $s$ and $t$ such that the first three terms are $a$, $b$, $c$.

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