# orthogonality of complex exponentials

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.

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.

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. $$OE\Delta$$.