In the question in the link is it compulsory that $A+B+C=\pi$ ?
Answer
Here in Problem when $x=\cos (a) +i\sin(a),\ y=\cos (b) +i\sin(b),\ z=\cos (c) +i\sin(c),\ x+y+z=0$,
we have proved that $$\sum\cos A=\sum\sin A=0\ \ \ \ (1)$$ $$(1)\implies\sum\cos2A=\sum\sin2A=0\ \ \ \ (2)$$
Now from $(1),\cos A+\cos B=-\cos C$ and $\sin A+\sin B=-\sin C$
Squaring & adding we get, $\cos(A-B)=-\dfrac12\ \ \ \ (3)$
Similarly, $\cos(B-C)=-\dfrac12\ \ \ \ (4),\cos(C-A)=-\dfrac12\ \ \ \ (5)$
$(3)\implies A-B=2m\pi\pm\dfrac{2\pi}3$ where $m$ is any integer
$(4)\implies B-C=2n\pi\pm\dfrac{2\pi}3$ where $n$ is any integer
If we take the opposite signs, $A-C=A-B+B-C=2\pi(m-n)$
$\implies\cos(C-A)=\cos(A-C)=\cos2\pi(m-n)=1$ which contradicts $(5)$
So, either $A-B=2m\pi+\dfrac{2\pi}3,B-C=2n\pi+\dfrac{2\pi}3\ \ \ (6)$
Or $A-B=2m\pi-\dfrac{2\pi}3,B-C=2n\pi-\dfrac{2\pi}3\ \ \ (7)$
Observe that both$(6),(7)$ satisfy $(5)$
So, the angles have to differ by $\dfrac{2\pi}3\pmod{2\pi}$ which is the sufficient condition for $(1)$ which always implies $(2)$
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