If the equation $z^4+a_1{z}^3+a_2{z}^2+a_{3}z+a_4=0$ $\left(wherea,b,c\in R\right)$has a purely imaginary root, then $\frac{a_3}{a_1{a}_2}+\frac{a_1{a}_4}{a_2{a}_3}=?$

Let $\lambda i$ be the root $\left(\lambda \ne 0\right).$ put $\lambda i$ in given equatioin

$\left({\lambda }^4-a_2{\lambda }^2+a_4\right)+i\left(a_3\lambda -a_1{\lambda }^3\right)=0$

$\Rightarrow {\lambda }^4-a_2{\lambda }^2+a_4=0-----\left(1\right)$

$a_3\lambda -a_1{\lambda }^3=0----\left(2\right)$

Now(2) $\Rightarrow {\lambda }^2=\frac{a_3}{a_1}.$ substitute ${\lambda }^2=\frac{a_3}{a_1}in\left(1\right)gives{a}_3^2+a_4{a}_1^2=a_1{a}_2{a}_3$

Divide with $a_1{a}_2{a}_3\Rightarrow \frac{a_3}{a_1{a}_2}+\frac{a_4{a}_1}{a_2{a}_3}=1$