# Show that $w_3=\frac{-1+i\sqrt3}2$ without $\cos$ and $\sin$

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 $$w_3^2 + w_3 + 1 = 0$$.

• 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 $x^3-1=(x-1)(x^2+x+1)=0$ Nov 26 '20 at 4:55
• Actually sounds very good, thanks ! Nov 26 '20 at 10:57

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 $$w^3=w^2w=1$$.)
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 $$x^2+x+1=0$$.