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 $\displaystyle\arg(z^2-z) =[-\pi,\frac{\pi}{3}]$ ???
 A: 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 
A: Here my extended hint for $2$:
Notice that $z^2-z = z(z-1) = e^{i\alpha}(e^{i\alpha}-1)$. And since $e^{i(\alpha+\beta)} = e^{i\alpha} e^{i\beta}$ therefore $\arg(ab)=\arg(a)+\arg(b)$
So $\arg(z^2-z) = \alpha + \arg(e^{i\alpha}-1)$
You will now find geometrically that $\arg(e^{i\alpha}-1) = \pi/2 + \alpha/2$ (I hope this is correct.)
You just have to consider the triangle $(0,e^{i\alpha},1)$
A: Let:
$z_{1} = a + ib \\ z_2 =p+iq \\ z_{3} = x + iy$
Taking the square of modulus:
$a^ 2 +b^ 2 =p^ 2 +q^ 2 =x^ 2 +y^ 2 =1$
Equating the real and the imaginary part of $z_1 + z_2 + z_3$ to 0,
$a + p + x = b + q + y = 0$
$
|z_{1} - z_{2}|^ 2 \\
=(a - p) ^ 2 + (b - q) ^ 2 \\
=2(a^ 2 + p^ 2 )-(a+p)^ 2 + 2(b ^ 2 + q ^ 2) - (b + q) ^ 2 \\ 
=2(a^ 2 + b^ 2 + p ^ 2 + q ^ 2 ) - x ^ 2 - y ^ 2\\
=4 - 1
=3
$
Similarly, you can show all sides are equal.
A: A more visual approach to $1)$, using roots of unity as OP requested:
If we consider the numbers as vectors in the plane, the conditions $|z_i|=1$ and $z_1+z_2+z_3=0$ are shown in Figure 1.

$\tag{Fig. 1}$
Let $u$ be any complex number such that $|u|=1$, then the transformation of the plane $z \mapsto uz$ is a rotation of the plane - and thus also an isometry.  This means $$|uz-uw|=|z-w|,$$ for any two complex numbers $z$ and $w$. In particular, consider the transformation which maps $z_2$ to $\omega=e^{\frac{i2\pi}{3}}$ as shown in Figure 2.

$\tag{Fig. 2}$
Since we know the roots of unity form an equilateral triangle $\triangle_{\omega}$ in the plane, we need only show $uz_1$ and $uz_3$ are the other two roots of unity and we're done. To see this, consider the centroid $G_u$ of the triangle $\triangle_u$ in Figure 2. This is the vector average $$\frac{1}{3}(uz_1+uz_2+uz_3)=\frac{1}{3}u(z_1+z_2+z_3)=0$$ and hence $G_u=G_{\omega}$. Now, $G_u=G_{\omega}$ is the point $2/3$rds of the way from $uz_2=\omega$, on the line $L$ passing through the midpoints between $uz_1$ and $uz_3$ and $\omega^2$ and $1$, respectively - and since only one chord on the unit circle can satisfy these conditions simultaneously, the result follows.
$\square$
A: 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.
