Let $ a,b$ be two complex numbers. Prove that roots of $ z^{4}+az^{2}+b$ form a rhombus with origin as center, if and only if $ \frac{a^{2}}{b}$ is a non-positive real number.
we suppsoe that roots of $ z^{4}+az^{2}+b$ form a rhombus with origin as center
it's evident that $ \exists (u,v,w)\in\mathbb{R}^{3},\forall z\in C:\ z^{4}+az^{2}+b=(z^{2}-u^{2}e^{iw})(z^{2}-v^{2}e^{iw})$
then $ a=-e^{iw}(u^{2}+v^{2})$ and $ b=(uv)^{2}e^{i(2w)}$ so $ \frac{a^{2}}{b}=\frac{(u^{2}+v^{2})^{2}}{(uv)^{2}}\in R^{+}$ non-negative real number.
Omid Hatami wrote:
Let $ a,b$ be two complex numbers. Prove that roots of $ z^{4} + az^{2} + b$ form a rhombus
with origin as center, if and only if $ \frac {a^{2}}{b}$ is a non-positive real number.
Proof. The roots of the given equation form a rhombus $ \Longleftrightarrow$
exist $ w\in\mathcal C$ and $ r>0$ so that these roots are $ \pm w$ and $ \pm riw$ .
Thus, $ z^4+az^2+b=(z^2-w^2)(z^2+r^2w^2)$ $ \Longleftrightarrow$
$ a=(r^2-1)w^2$ and $ b=-r^2w^4$ $ \Longrightarrow$ $ \frac {a^2}{b}=-\left(\frac {r^2-1}{r}\right)^2\le 0\ .$