Show that complex numbers are vertices of equilateral triangle 1)Show if $|z_1|=|z_2|=|z_3|=1$ and $z_1+z_2+z_3=0$ then $z_1,z_2,z_3$ are vertices of equilateral triangle inscribed in a circle of radius.
I thought I can take use from roots of unity here, since $|z_1|=|z_2|=|z_3|=1$ they lie at circle at radius $1$ but I don't know how to take advantage from $z_1+z_2+z_3=0$
2)Let $z=\cos\alpha+i\sin\alpha$ where $\alpha \in 0,2\pi$ then find $\arg(z^2-z)$
I come to this siutation $\displaystyle z^2-z=-2\sin{\frac{1}{2}x}(\sin{\frac{3}{2}x}+i\cos{\frac{3}{2}x})=-2\sin{\frac{1}{2}x}(\cos(\frac{\pi}{2}-{\frac{3}{2}x})+i\sin({\frac{\pi}{2}-\frac{3}{2}x}))$ so $\displaystyle 0\le\frac{\pi}{2}-\frac{3}{2}x\le2\pi$ so $\displaystyle\frac{\pi}{3}\ge x \ge - \pi$ so $\displaystyle\arg(z^2-z) =[-\pi,\frac{\pi}{3}]$ ???

Here is a simple way. Let $z_{1}=cos\theta_{1}+isin\theta_{1}$,$z_{2}=cos\theta_{2}+isin\theta_{2}$, $z_{3}=cos\theta_{3}+isin\theta_{3}.$
By $z_{1}+z_{2}+z_{3}=0$ we get the sum of the cosines are zero and the sum of the sines is zero.
Squaring the equations
$cos\theta_{1}+cos\theta_{2}=-cos\theta_{3}$ and the
$sin\theta_{1}+sin\theta_{2}=-sin\theta_{3}$ we get
$cos(\theta_{1}-\theta_{2})=-\dfrac{1}{2}$.
Therefore $\theta_{1}-\theta_{2}=\dfrac{2\pi}{3}$. Likewise we get similar equations.
Since the triangle with sides $|z_{1}|$ and $|z_{2}|$ is isosceles, and
the angle $(z_{1},z_{2})$ is $120^{o}$ then the other two angles are $30^{o}$.
We do the same for $z_{2},z_{3}$ and we get the same result.
So the angles of the triangle are all $30^{0}+30^{0}=60^{0}$ and hence the triangle is equilateral.
