Condition for Quartic with Real Coefficients to have Wholly Imaginary Root

Theorem
Let $Q$ be the quartic equation:
 * $(1): \quad z^4 + a_1 z^3 + a_2 z^2 + a_3 z + a_4 = 0$

such that all of $a_1, a_2, a_3, a_4$ are real numbers.

Then $Q$ has a root which is wholly imaginary :


 * $\text {(a)}: \quad a_3^2 + a_1^2 a_4 = a_1 a_2 a_3$
 * $\text {(b)}: \quad a_1 a_3 > 0$

Necessary Condition
We have:

This leads to the factorisation of $(1)$:

Two solutions to this are found from:

The other two solutions are found by solving:

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:

Hence:

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.