condition on complex numbers to form cyclic quadrilateral. Consider complex numbers $z,z^2,z^3,z^4$  in that order which form a cyclic quadrilateral . If $\arg z=\alpha$ and $\alpha$ lies in $[0,2\pi]$.Find the values $\alpha$ can take.
I encountered this question in one competitive exam.i tried using the property of cyclic quadrilateral to get $$\arg\left(\frac{z^3-z^4}{z-z^4}\right)+\arg\left(\frac{z-z^2}{z^3-z^2}\right)=\pi$$ This can be simplfied furher but does not help.
I also tried using coni theorem but of no use. Answer given is alpha lies in $(0,\frac{2\pi}{3})and(\frac{4\pi}{3},2\pi)$

By Ptolemy we obtain:
$$|z-z^2|\cdot|z^3-z^4|+|z-z^4|\cdot|z^2-z^3|=|z-z^3|\cdot|z^2-z^4|$$ or
$$|z|+|z^2+z+1|=|(z+1)^2|.$$
Now, we can use a triangle inequality.
Id est, for $|z|=r$ we obtain:
$$(\cos\alpha,\sin\alpha)||(r^2\cos2\alpha+r\cos\alpha+1,r^2\sin2\alpha+r\sin\alpha),$$ which gives
$$\sin\alpha(r^2\cos2\alpha+r\cos\alpha+1)=\cos\alpha(r^2\sin2\alpha+r\sin\alpha)$$ or
$$\sin\alpha=r^2\sin\alpha$$ and since $\sin\alpha\neq0$, we obtain $r=1$.
