Prove if sum of complex numbers = 0 and their magnitudes = 1, then sum of their squares = 0 This problem is from Chapter 6 (Basics of Complex Numbers) of the AoPS Precalculus textbook.
Show that if complex numbers $ w_1+w_2+w_3 = 0$ and $|w_1|=|w_2|=|w_3| = 1$, then $w_1^2 + w_2^2 + w_3^2 = 0$.
I really didn't know what to do here. My approach was to get from the sum of squares equation to the first equation. I tried multiplying the first equation by $\overline {w_1w_2w_3}$ so I could introduce magnitudes, and substitute those for $1$, but after that I was just left with $w_1\overline{w_2w_3} + w_2\overline{w_1w_3} + w_3\overline{w_1w_2} = 0$. I couldn't get anywhere from there.
Any help would appreciated, thank you.
 A: The first step is to understand what the input means. $|w_1|=|w_2|=|w_3|=1$ means that you have three points on the unit circle in the complex plane. $w_1+w_2+w_3=0$ means that the centroid of the triangle in the complex plane formed by these three points is at the origin. It can be shown that the two conditions mean that the three points form an equilateral triangle.
We can then use the exponential notation for complex points. $$w_1=e^{i\phi}\\w_2=e^{i(\phi+2\pi/3)}\\w_3=e^{i(\phi+4\pi/3)}$$
We write the first equation as
$$e^{i\phi}+e^{i(\phi+2\pi/3)}+e^{i(\phi+4\pi/3)}=e^{i\phi}(1+e^{i2\pi/3}+e^{i4\pi/3})=0$$
Now let's square all the rems and add them:
$$w_1^2+w_2^2+w_3^2=e^{2i\phi}+e^{2i(\phi+2\pi/3)}+e^{2i(\phi+4\pi/3)}\\=e^{2i\phi}(1+e^{2i2\pi/3}+e^{2i4\pi/3})\\=e^{2i\phi}(1+e^{i4\pi/3}+e^{i8\pi/3})\\=e^{2i\phi}(1+e^{i4\pi/3}+e^{i2\pi/3})=0$$
I've used here that $e^{i6\pi/3}=e^{i2\pi}=1$.
EDIT: Doing the same thing without exponential form.
Since $|w_1|=1$, we know that $w_1\ne 0$. Let's divide your fist equation by $w_1$:
$$1+\frac{w_2}{w_1}+\frac{w_3}{w_1}=0$$
We also use $$\left|\frac{w_2}{w_1}\right|=\frac{|w_2|}{|w_1|}=1$$ and similar for $w_3$. Then we can use $v_2=\frac{w_2}{w_1}$ and $v_3=\frac{w_3}{w_1}$, with $|v_2|=|v_3|=1$. So your problem can be rewritten as: given $|v_2|=|v_3|=1$ and $1+v_2+v_3=0$, show $1+v_2^2+v_3^2=0$.
Since the absolute values are $1$, we can write $$v_2=\cos\alpha+i\sin\alpha\\v_3=\cos\beta+i\sin\beta$$
Then writing by components we get $$1+\cos\alpha+\cos\beta=0\\\sin\alpha+\sin\beta=0$$
From the last equation you get $\beta=-\alpha$. Then $$1+\cos\alpha+\cos(-\alpha)=1+2\cos\alpha=0\\\alpha=\frac{2\pi}3\\\beta=-\frac{2\pi}3$$
These last two equations means that $1+0i, v_2, v_3$ are forming an equilateral triangle in the complex plane. It should be easy now to calculate $v_2^2$ and $v_3^2$.
$$v_2^2=(\cos\alpha+i\sin\alpha)^2=\cos^2\alpha+\sin^2\alpha+2i\cos\alpha\sin\alpha=\cos(2\alpha)+i\sin(2\alpha)$$
Similarly, $$v_3^2=\cos(2\alpha)-i\sin(2\alpha)$$
Then $$1+v_2^2+v_3^2=1+2\cos(2\alpha)=1+2\cos\frac{4\pi}3=1+2\left(-\frac12\right)=0$$
A: If you multiply the equation $\omega_1+\omega_2+\omega_3=0$ by conjugate($\omega_1\omega_2\omega_3$) then you arrive at conjugate ($\omega_2\omega_3$) $+$ conjugate ($\omega_1\omega_3$) $+$ conjugate ($\omega_1\omega_2$)$=0$ therefore $\omega_2\omega_3+\omega_1\omega_3+\omega_1\omega_2=0$
Since $(\omega_1+\omega_2+\omega_3)^2 = \omega_1^2+ \omega_2^2+ \omega_3^2 + 2( \omega_2\omega_3+\omega_1\omega_3+\omega_1\omega_2)$ it follows immediately that   $\omega_1^2+ \omega_2^2 +\omega_3^2=0$
