# 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 if and only if:

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

## Proof

### Necessary Condition

We have:

 $\displaystyle a_3^2 + a_1^2 a_4$ $=$ $\displaystyle a_1 a_2 a_3$ $\displaystyle \leadsto \ \$ $\displaystyle a_4$ $=$ $\displaystyle \frac {a_3} {a_1} \paren {a_2 - \frac {a_3} {a_1} }$

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

 $\displaystyle z^4 + a_1 z^3 + a_2 z^2 + a_3 z + a_4$ $=$ $\displaystyle 0$ $\displaystyle \leadsto \ \$ $\displaystyle \paren {z^2 + \frac {a_3} {a_1} } \paren {z^2 + a_1 z + \paren {a_2 - \frac {a_3} {a_1} } }$ $=$ $\displaystyle 0$

Two solutions to this are found from:

 $\displaystyle z^2 + \frac {a_3} {a_1}$ $=$ $\displaystyle 0$ $\displaystyle \leadsto \ \$ $\displaystyle z^2$ $=$ $\displaystyle -\frac {a_3} {a_1}$ $\text {(2)}: \quad$ $\displaystyle \leadsto \ \$ $\displaystyle z$ $=$ $\displaystyle \pm \sqrt {-\frac {a_3} {a_1} }$

The other two solutions are found by solving:

 $\displaystyle z^2 + a_1 z + \paren {a_2 - \frac {a_3} {a_1} }$ $=$ $\displaystyle 0$ $\text {(3)}: \quad$ $\displaystyle \leadsto \ \$ $\displaystyle z$ $=$ $\displaystyle -\frac {a_1} 2 \pm \frac {\sqrt {a_1^2 - 4 \paren {a_2 - a_3 / a_1} } } 2$ Quadratic Formula

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:

 $\text {(4)}: \quad$ $\displaystyle a_3 - a_1 b^2$ $=$ $\displaystyle 0$ $\text {(5)}: \quad$ $\displaystyle a_4 - \paren {a_2 - b^2} b^2$ $=$ $\displaystyle 0$

Hence:

 $\displaystyle b^4 - a_2 b^2 + a^4$ $=$ $\displaystyle 0$ from $(4)$ $\displaystyle \leadsto \ \$ $\displaystyle b^2$ $=$ $\displaystyle \frac {a_2 \pm \sqrt {a_2^2 - 4 a^4} } 2$ Quadratic Formula $\text {(6)}: \quad$ $\displaystyle \leadsto \ \$ $\displaystyle \paren {2 b^2 - a_2}^2$ $=$ $\displaystyle a_2^2 - 4 a^4$ rearranging and squaring both sides $\displaystyle b^2$ $=$ $\displaystyle \frac {a_3} {a_1}$ from $(5)$ $\displaystyle \leadsto \ \$ $\displaystyle \paren {\frac {2 a_3} {a_1} - a_2}^2$ $=$ $\displaystyle a_2^2 - 4 a^4$ substituting for $b^2$ into $(6)$, rearranging and squaring both sides $\displaystyle \leadsto \ \$ $\displaystyle \paren {\frac {2 a_3} {a_1} - a_2}^2$ $=$ $\displaystyle a_2^2 - 4 a^4$ substituting for $b^2$ into $(6)$, rearranging and squaring both sides $\displaystyle \leadsto \ \$ $\displaystyle 4 \frac {a_3^2} {a_1^2} + a_2^2 - \frac {4 a_2 a_3} {a_1}$ $=$ $\displaystyle a_2^2 - 4 a_4$ $\displaystyle \leadsto \ \$ $\displaystyle a_3^2 + a_1^2 a_4$ $=$ $\displaystyle a_1 a_2 a_3$

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.

$\blacksquare$

## Sources

(although see Condition for Quartic with Real Coefficients to have Wholly Imaginary Root/Mistake for analysis of an error in that work)