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I struggle a lot to show that $w_3 = e^{i2\pi/3}$ is equal to $\frac{-1+i\sqrt{3}}{2}$ without using cos and sin, knowing only that $f :(\mathbb R, +) \to (\mathbb C(1), \times)$ that from an angle $\theta$ gives the complex number $e^{i\theta}$ is a morphism so $\operatorname{ker} f = 2\pi \mathbb Z$, also knowing $\cos(\theta) = \Re(e^{i\theta})$, $\sin(\theta) = \Im(e^{i\theta})$, knowing $e^{i\pi}=-1$ all summation and duplication rules of $\cos$ and $\sin$. The indication tells to verify that $w_3$ is a solution of $X^{2} + X + 1 = 0$ then compute the complex solutions with $\Delta$ and find that since $w_3$ is a solution so it must be one of them, without knowing what $\cos(\pi/3)$ is. Tried all possible methods but still need to find that value of $\cos(\pi/3)$ but the whole goal of the exercice is to do it without knowing it. I also wrote $f(\pi) = f(2\pi/3 + \pi/3) = f(2\pi/3)\cdot f(\pi/3)$ but leads me nowhere... I would like your help to show that $w_3^2 + w_3 + 1 = 0$.

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    $\begingroup$ I’m not sure I understand what you’re asking, but you could solve $x^2+x+1=0$ using the quadratic equation, and then $x^3-1=(x-1)(x^2+x+1)=0$ $\endgroup$ Nov 26 '20 at 4:55
  • $\begingroup$ Actually sounds very good, thanks ! $\endgroup$
    – Chady
    Nov 26 '20 at 10:57
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show that $w = e^{i2\pi/3}$ is equal to $\dfrac{-1+i\sqrt3}{2}$ without using cos and sin

Well, $w^3=1$ and $w\ne1$, so $\dfrac{w^3-1}{w-1}=w^2+w+1=0.$

Now, by the quadratic formula, $w=\dfrac{-1\pm\sqrt{-3}}2.$

(Note that $w^2=\dfrac{-1\mp\sqrt{-3}}2$, and you can verify that $w^3=w^2w=1$.)

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When viewed as vectors in the complex plane, $1,w_3,w_3^2$ are equally spaced around the origin and are of the same length. Thus their sum must be zero and we have $w_3$ as a solution to $x^2+x+1=0$.

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