Condition for Quartic with Real Coefficients to have Wholly Imaginary Root

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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:

\(\ds a_3^2 + a_1^2 a_4\) \(=\) \(\ds a_1 a_2 a_3\)
\(\ds \leadsto \ \ \) \(\ds a_4\) \(=\) \(\ds \frac {a_3} {a_1} \paren {a_2 - \frac {a_3} {a_1} }\)

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

\(\ds z^4 + a_1 z^3 + a_2 z^2 + a_3 z + a_4\) \(=\) \(\ds 0\)
\(\ds \leadsto \ \ \) \(\ds \paren {z^2 + \frac {a_3} {a_1} } \paren {z^2 + a_1 z + \paren {a_2 - \frac {a_3} {a_1} } }\) \(=\) \(\ds 0\)


Two solutions to this are found from:

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


The other two solutions are found by solving:

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

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


Hence:

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


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)
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