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I want to show that $(1+e^{-j2/3 \pi}+e^{-j4/3 \pi}) = 0$.

The property that I am supposed to use is that of orthogonality of complex exponentials. This means that if $e^{-j(2\pi kn)/N }$ will be equal to N when $n=lN$.

My problems are that the rule is applied when its a sum and this case its not a sum. Secondly, there is no $ln$ in those exponents such that the properties are achieved. Finally even if we apply the property it would give N which is $3$ in this case so I do not see how it would add up to $0$.

Thank you

edit:

This is is the orthogonality of complex exponentials.

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Assuming you meant orthogonality of characters of $\mathbb{Z}/N\mathbb{Z}$ $$\sum_{x=0}^{N-1}\chi_k(x)\chi_{k'}(x)^{-1}= \begin{cases}N & k = k'\\ 0 & k \neq k' \end{cases}$$ where $\chi_k(x)=\mathrm{e}^{2\pi\mathrm{i}k x/N}$.

Then consider $N=3$, $k=0$, $k'=1$.

To actually prove orthogonality, use the argument with cyclotomic polynomials in the other answer.

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Not sure what is meant here by "orthogonality of complex exponentials"; also, I'm having a little difficulty with the assertion that "$e^{-j(2\pi kn)/N}$ will be equal to $N$ when $n = lN$," since

$\vert e^{-2j\pi kn)/N} \vert = 1, \tag 1$

always, whereas $N$ with

$\vert N \vert \ge 2 \tag 2$

appear admissible in the context of the question itself; thus the assertion of equality here seems somewhat murky.

Despite these and other semantic difficulties, that

$1 + e^{-2j\pi /3} + e^{-4j \pi /3} = 0 \tag 3$

is easily seen; setting

$\omega = e^{-2j \pi/ 3} \tag 4$

for brevity, we have

$\omega^2 = e^{-4j \pi/ 3} \tag 5$

and

$\omega^3 = e^{-6j \pi / 3} = e^{-2j \pi} = 1; \tag 6$

thus

$(\omega - 1)(\omega^2 + \omega + 1) = \omega^3 - 1 = 0; \tag 7$

now since (4) implies

$\omega - 1 = e^{-2j \pi/ 3} - 1 \ne 0, \tag 8$

(7) yields

$1 + \omega + \omega^2 = 0, \tag 9$

when we substitute (4) and (5) into this equation we are left with (3), the desired result. $OE\Delta$.

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