=== Necessary Condition === We have: {{begin-eqn}} {{eqn | l = a_3^2 + a_1^2 a_4       | r = a_1 a_2 a_3       | c =  }} {{eqn | ll= \leadsto       | l = a_4       | r = \frac {a_3} {a_1} \paren {a_2 - \frac {a_3} {a_1} }       | c =  }} {{end-eqn}} This leads to the factorisation of $(1)$: {{begin-eqn}} {{eqn | l = z^4 + a_1 z^3 + a_2 z^2 + a_3 z + a_4       | r = 0       | c =  }} {{eqn | ll= \leadsto       | l = \paren {z^2 + \frac {a_3} {a_1} } \paren {z^2 + a_1 z + \paren {a_2 - \frac {a_3} {a_1} } }       | r = 0       | c =  }} {{end-eqn}} Two solutions to this are found from: {{begin-eqn}} {{eqn | l = z^2 + \frac {a_3} {a_1}       | r = 0       | c =  }} {{eqn | ll= \leadsto       | l = z^2       | r = -\frac {a_3} {a_1}       | c =  }} {{eqn | n = 2       | ll= \leadsto       | l = z       | r = \pm \sqrt {-\frac {a_3} {a_1} }       | c = }} {{end-eqn}} The other two solutions are found by solving: {{begin-eqn}} {{eqn | l = z^2 + a_1 z + \paren {a_2 - \frac {a_3} {a_1} }       | r = 0       | c =  }} {{eqn | n = 3       | ll= \leadsto       | l = z       | r = -\frac {a_1} 2 \pm \frac {\sqrt {a_1^2 - 4 \paren {a_2 - a_3 / a_1} } } 2       | c = Quadratic Formula }} {{end-eqn}} As $a_1 \ne 0$ it follows that $(3)$ always has a real part. So only the $(2)$ can produce roots which are wholly imaginary, and that can happen only when: : $a_3 / a_1 > 0$ that is: : $a_1 a_3 > 0$ Thus, if $a_3^2 + a_1^2 a_4 = a_1 a_2 a_3$ and $a_1 a_3 > 0$, $Q$ has wholly imaginary roots. === Sufficient Condition === Let $z^4 + a_1 z^3 + a_2 z^2 + a_3 z + a_4 = 0$ have wholly imaginary roots. Then it can be written: :$\paren {z^2 + b^2} \paren {z^2 + m z + n} = 0$ By long division of $Q$ by $z^2 + b^2$: :$\paren {z^2 + b^2} \paren {z^2 + a_1 z + \paren {a_2 + b^2} } + \paren {a_3 - a_1 b^2} z + a_4 - \paren {a_2 - b^2} b^2 = 0$ For this to be in the above format, it is necessary that: :$\paren {a_3 - a_1 b^2} z + a_4 - \paren {a_2 - b^2} b^2 \equiv 0$ That is: {{begin-eqn}} {{eqn | n = 4       | l = a_3 - a_1 b^2       | r = 0       | c =  }} {{eqn | n = 5       | l = a_4 - \paren {a_2 - b^2} b^2       | r = 0       | c =  }} {{end-eqn}} Hence: {{begin-eqn}} {{eqn | l = b^4 - a_2 b^2 + a^4       | r = 0       | c = from $(4)$ }} {{eqn | ll= \leadsto       | l = b^2        | r = \frac {a_2 \pm \sqrt {a_2^2 - 4 a^4} } 2       | c = Quadratic Formula }} {{eqn | n = 6       | ll= \leadsto       | l = \paren {2 b^2 - a_2}^2        | r = a_2^2 - 4 a^4       | c = rearranging and squaring both sides }} {{eqn | l = b^2        | r = \frac {a_3} {a_1}       | c = from $(5)$ }} {{eqn | ll= \leadsto       | l = \paren {\frac {2 a_3} {a_1} - a_2}^2        | r = a_2^2 - 4 a^4       | c = substituting for $b^2$ into $(6)$, rearranging and squaring both sides }} {{eqn | ll= \leadsto       | l = \paren {\frac {2 a_3} {a_1} - a_2}^2        | r = a_2^2 - 4 a^4       | c = substituting for $b^2$ into $(6)$, rearranging and squaring both sides }} {{eqn | ll= \leadsto       | l = 4 \frac {a_3^2} {a_1^2} + a_2^2 - \frac {4 a_2 a_3} {a_1}       | r = a_2^2 - 4 a_4       | c =  }} {{eqn | ll= \leadsto       | l = a_3^2 + a_1^2 a_4       | r = a_1 a_2 a_3       | c =  }} {{end-eqn}} Hence if $z^4 + a_1 z^3 + a_2 z^2 + a_3 z + a_4 = 0$ has wholly imaginary roots, then: :$a_3^2 + a_1^2 a_4 = a_1 a_2 a_3$ We also see that the roots themselves are $\pm b$, where: :$b = \sqrt {-\dfrac {a_3} {a_1}}$ Thus we have also shown that $a_3 a_1 > 0$ in order for those roots to indeed be wholly imaginary as required. {{qed}}