I have to prove that $z_1, z_2, z_3, z_4$, where $|z_1| = |z_2| = |z_3| = |z_4| = 1$, are the vertices of a rectangle if and only if $z_1z_2z_3+z_1z_2z_4+z_1z_3z_4+z_2z_3z_4=0$
Any help?
There is a hint in my book about the converse, that says that since this equity holds, then 1)the polynomial with roots $z_i$ is of the form $p(z)=z^4+az^2+b$, 2) hence its four roots are devised in two pairs of opposite numbers. I do not understand why both of these hold
Let one vertex $z_1$ be $e^{j\theta_1}$ and second vertex $z_2$be $e^{j\theta_2}$, then third vertex $z_3$ is diametrically opposite to $V_1$, as $\angle z_1z_2z_3 = 90^{\circ}$, so $z_3 = e^{j\theta_1+\pi}$ and similarly, $z_4=e^{j\theta_2+\pi}$. Substituting in $z_1z_2z_3+z_1z_2z_4+z_1z_3z_4+z_2z_3z_4 = e^{j(\theta_1+\theta_2+\theta_1+\pi)} + e^{j(\theta_2+\theta_1+\pi+\theta_2+\pi)}+e^{j(\theta_1+\pi+\theta_2+\pi+\theta_1)}+e^{j(\theta_2+\pi+\theta_1+\theta_2)}$ First and third terms and second and fourth terms cancel to give zero. [edit]: Attempt for converse: $z_1z_2z_3+z_1z_2z_4+z_1z_3z_4+z_2z_3z_4=0$ clearly, $z_1z_2z_3z_4 \ne 0$, so $\frac{1}{z_1}+\frac{1}{z_2}+\frac{1}{z_3}+\frac{1}{z_4} = 0$, but $\frac{1}{z_i}=e^{-j\theta_i}$, so $z_1+z_2+z_3+z_4 = \sum e^{j\theta_i} = 0$ That is expression with factors $x-z_i$ is such that odd terms are zero, and is of form $P(x)=x^4+ax^2+b$ (since sum of roots and sum of inverses of roots are both zero). And so the roots of $P(x)=0$ are of form $\pm e^{\pm j\theta}$ which can be written as $e^{j\theta}, e^{-j\theta}, e^{j(\theta+\pi)} and e^{-j(\theta+\pi)}$ Two pairs are ends of the diameters, so the resulting quadrilateral is a rectangle.
For first try write $$P(z)=(z-z_1)(z-z_2)(z-z_3)(z-z_4)$$ and open brackets. For second if $$P(z)=0$$, then $$P(-z)=P(z)=0.$$
