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 $\displaystyle\arg(z^2-z) =[-\pi,\frac{\pi}{3}]$ ???

Let: $z_1 =e^{ia} ; z_2 = e^{ib}; z_3 = e^{ic}$
$ z_1 +z_2 = e^{i\frac{a+b}{2}}*(e^{i\frac{a-b}{2}} + e^{-i\frac{(a-b)}{2}}) = e^{i\frac{a+b}{2}}*2*cos(\frac{a-b}{2}) = -z_3 $
=> $|2*cos(\frac{a-b}{2})| = |-z_3| = |z_3| = 1$ , 
If $ cos(\frac{a-b}{2}) =\frac{1}{2} $ -> $a = b \pm \frac{2\pi}{3}$ $mod(2\pi)$
here without loss of generality you can assume a= b+ $\frac{2\pi}{3}$ $ mod(2\pi)$ (the other case is the same)
you get : $\frac{a+b}{2} = c+\pi$  $ mod(2\pi)$  -> b+ $\frac{\pi}{3} = c + \pi$ $ mod(2\pi)$ -> $ b = c + \frac{2\pi}{3} $  $ mod(2\pi)$
You get your equilateral triangle, since you proved that you can rotate of $\frac{2\pi}{3}$ to pass from one point to another. The other cases are exactly the same. 
As for 2) , I would use : $z= e^{ia}$
$z^2 - z = e^{2ia} - e^{ia}$ = $e^{\frac{3}{2}ia}*2i*sin(\frac{a}{2})  $ = $ e^{(\frac{3}{2}a + \frac{\pi}{2})i}*2*sin(\frac{a}{2}) $. The sign of the sin is the only thing you have take into account to evaluate correctly the argument. If it is negative, you add $\pi$, else you already have your argument 
