Finding out the range of values of the argument of a complex number Let $z,z^2,z^3,z^4$ be the four complex numbers.If these taken in order form a cyclic quadrilateral then the question  is to find out the range of values of $\theta$ where $\theta$=$arg(z)$ and $\theta €(0,2\pi)$.
Since it forms a cyclic quadrilateral i applied coni's theorem to get $$\frac{(z^3-z^2)(z-z^4)}{(z-z^2)(z^3-z^4)}$$  as a purely real number.I simplified to get $z+1/z$ as a real number and from this i guessed $ |z|=1$.But i could proceed after this..Please help me in this regard.Thanks.
 A: You've already got $|z|=1$ correctly.
Note here that we have to have
$$z\not=z^2,z\not=z^3,z\not=z^4,z^2\not=z^3,z^2\not=z^4,z^3\not=z^4$$
giving 
$$\theta\not=\frac{2}{3}\pi,\pi,\frac{4}{3}\pi$$
Also, we have to consider the expression "these taken in order".
For $0\lt\theta\le\frac{\pi}{2}$, since we have $0\lt\theta\lt 2\theta\lt 3\theta\lt 4\theta\le 2\pi$, these $\theta$ are sufficient.
For $\frac{\pi}{2}\lt\theta\lt\frac{2}{3}\pi$, since we have $\frac{\pi}{2}\lt\theta\lt 2\theta\lt 3\theta\lt 2\pi\lt 4\theta\lt 4\pi$ and $0\lt 4\theta-2\pi\lt \theta\lt 2\theta\lt 3\theta\lt 2\pi$, these $\theta$ are sufficient.
For $\frac{2}{3}\pi\lt\theta\lt\pi$, since we have $\frac{2}{3}\pi\lt \theta\lt 2\theta\lt 2\pi\lt 3\theta\lt 4\theta\lt 4\pi$ and $0\lt 3\theta-2\pi\lt\theta\lt 4\theta-2\pi\lt 2\theta\lt 2\pi$, these $\theta$ are not sufficient.
For $\pi\lt\theta\lt\frac{4}{3}\pi$, since we have $\pi\lt\theta\lt 2\pi\lt 2\theta\lt 3\theta\lt 4\pi\lt 4\theta\lt 6\pi$ and $0\lt 2\theta-2\pi\lt 4\theta-4\pi\lt \theta\lt 3\theta-2\pi\lt 2\pi$, these $\theta$ are not sufficient.
For $\frac{4}{3}\pi\lt\theta\lt\frac{3}{2}\pi$, since we have $\frac{4}{3}\pi\lt \theta\lt 2\pi\lt 2\theta\lt 4\pi\lt 3\theta\lt 4\theta\lt 6\pi$ and $0\lt 3\theta-4\pi\lt 2\theta-2\pi\lt \theta\lt 4\theta-4\pi\lt 2\pi$, these $\theta$ are sufficient.
For $\frac{3}{2}\pi\le\theta\lt 2\pi$, since we have $\frac{3}{2}\pi\le \theta\lt 2\pi\lt 2\theta\lt 4\pi\lt 3\theta\lt 6\pi\le 4\theta\lt 8\pi$ and $0\le 4\theta-6\pi\lt 3\theta-4\pi\lt 2\theta-2\pi\lt \theta\lt 2\pi$, these $\theta$ are sufficient.
It follows from these that the answer is
$$\color{red}{0\lt\theta\lt\frac{2}{3}\pi\quad\text{or}\quad \frac{4}{3}\pi\lt\theta\lt 2\pi}$$
