<s>[INST] 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 $\displaystyle\arg(z^2-z) =[-\pi,\frac{\pi}{3}]$ ???
 [/INST] 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.
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