If the equation, ${\mathrm{z}}^4+{\mathrm{a}}_1{\mathrm{z}}^3+{\mathrm{a}}_2{\mathrm{z}}^2+{\mathrm{a}}_3\mathrm{z}+{\mathrm{a}}_4=0,$ where a₁, a₂, a₃, a₄, real coefficients are
different from zero has a pure imaginary root then the expression $\frac{{\mathrm{a}}_3}{{\mathrm{a}}_1{\mathrm{a}}_2}+\frac{{\mathrm{a}}_1{\mathrm{a}}_4}{{\mathrm{a}}_2{\mathrm{a}}_3}$ has the
value equal to:

${\mathrm{z}}^4+{\mathrm{az}}^3+{\mathrm{bz}}^2+\mathrm{cz}+\mathrm{d}=0,\mathrm{a},\mathrm{b},\mathrm{c},\mathrm{d}\in \mathrm{R}$
It has purely imaginary root $\mathrm{z}+\overline{\mathrm{z}}=0\Rightarrow \mathrm{z}=-\overline{\mathrm{z}}$
Taking conjugate on b.s of (i)
$\begin{array}{r}\begin{array}{rl} & \Rightarrow {\overline{\mathrm{z}}}^4+\mathrm{a}{\overline{\mathrm{z}}}^3+\mathrm{b}{\overline{\mathrm{z}}}^2+\mathrm{c}\overline{\mathrm{z}}+\mathrm{d}=0\\ & \Rightarrow {\mathrm{z}}^4-{\mathrm{az}}^3+{\mathrm{bz}}^2-\mathrm{cz}+\mathrm{d}=0\end{array}\\ \begin{array}{rl} & \text{ (i) }+\left(\text{ ii): }2\left({\mathrm{z}}^4+{\mathrm{bz}}^2+\mathrm{d}\right)=0\right.\\ & \Rightarrow {\mathrm{z}}^4+{\mathrm{bz}}^2+\mathrm{d}=0\\ & \text{ (i) }-\left(\mathrm{ii}\right):\\ & \mathrm{z}\left({\mathrm{az}}^2+\mathrm{c}\right)=0\\ & \mathrm{z}=0,{\mathrm{z}}^2=\frac{-\mathrm{c}}{\mathrm{a}}\\ & \frac{{\mathrm{c}}^2}{{\mathrm{a}}^2}-\frac{\mathrm{bc}}{\mathrm{a}}+\mathrm{d}=0\\ & \Rightarrow {\mathrm{c}}^2-\mathrm{abc}+{\mathrm{a}}^2\mathrm{d}=0\\ & \Rightarrow {\mathrm{c}}^2+{\mathrm{a}}^2\mathrm{d}=\mathrm{abc}\\ & \Rightarrow \frac{{\mathrm{c}}^2}{\mathrm{abc}}+\frac{{\mathrm{a}}^2\mathrm{d}}{\mathrm{abc}}=1\\ & \Rightarrow \frac{\mathrm{c}}{\mathrm{ab}}+\frac{\mathrm{ad}}{\mathrm{bc}}=1\end{array}\end{array}$