Necessary and sufficient conditions for a quadratic polynomial with complex coefficients to have both roots with negative real parts

Question

Namely, when does the polynomial $z^2 + a z + b = 0$ with $a,b \in \mathbb{C}$ have both roots in the left half complex?

As the sum of the roots $z_1 + z_2 = -a$, it follows that if $\Re a < 0$ then one of $\Re z_1$ or $\Re z_2$ must be positive. So a necessary condition is $\Re a > 0$. What is a sufficient condition? (Please provide a proof for it)

Edit: We can modify the answer below for the "more general case". A necessary and sufficient condition for the roots of $$ A z^2 + B z + C = 0$$ with $A>0$, $B,C \in \mathbb{C}$ to lie on the left complex half plane, i.e. $\Re z_i < 0$, is to have $$\sqrt{2} \Re(B) > \sqrt{\Re(B^2 - 4AC) + \vert B^2 - 4AC \vert}.$$ If we let $B = B_1 + i B_2$, $C = C_1 + i C_2$, with $B_i, C_i \in \mathbb{R}$ then the above condition can also be written as $$ B_1 > 0, \qquad B_1^2 C_1 + B_1 B_2 C_2 - A C_2^2 > 0. $$

Answer 1

If

$\mu_1, \mu_2 \in \Bbb C \tag 1$

are the roots of

$z^2 + az + b = 0, \; a, b \in \Bbb C, \tag 2$

then they satisfy the quadratic formula,

$\mu_1, \mu_2 = \dfrac{-a \pm \sqrt{a^2 - 4b}}{2}, \tag 3$

provided we take some care in the interpretation of $\sqrt{\cdot}$; since in general $a^2 - 4b \in \Bbb C \setminus \Bbb R$, we must carefully specify just what we intend

$\theta = \arg{\sqrt{a^2 - 4b}} \in [0, 2\pi) \tag 4$

to be; I think the most generally accepted convention, assuming $a^2 - 4b \ne 0$, is to write

$a^2 - 4b = \vert a^2 - 4b \vert e^{i \arg(a^2 - 4b)}, \tag 5$

and then choose (or, define)

$\sqrt{a^2 - 4b} = \sqrt{\vert a^2 - 4b \vert} e^{i \arg(a^4 - 4b) / 2}; \tag 6$

we check:

$(\sqrt{\vert a^2 - 4b \vert} e^{i \arg(a^2 - 4b) / 2})^2 = (\sqrt{ \vert a^2 - 4b \vert })^2 (e^{i \arg(a^4 - 4b) / 2})^2$ $= \vert a^2 - 4b \vert e^{2i\arg((a^2 - 4b) / 2)} = \vert a^2 - 4b \vert e^{i\arg(a^2 - 4b)} = a^2 - 4b; \tag 7$

of course with (6) in place, we also see that

$(-\sqrt{a^2 - 4b})^2 = a^2 - 4b \tag 8$

as well; and of course, (3) is a necessary for $\mu_1$, $\mu_2$ to satisfy (2), since it follows from completing the square: from (2),

$z^2 + az = -b, \tag 9$

$\left (z + \dfrac{a}{2} \right )^2 = z^2 + az + \dfrac{a^2}{4} = \dfrac{a^2}{4} - b = \dfrac{1}{4}(a^2 - 4b), \tag{10}$

whence

$z + \dfrac{a}{2} = \pm \dfrac{1}{2}\sqrt{a^2 - 4b} \iff \mu_1, \mu_2 = z = \dfrac{-a \pm \sqrt{a^2 - 4b}}{2}. \tag{11}$

It should be noted here that

$a^2 - 4b = 0 \Longrightarrow \mu_1 = \mu_2 = -\dfrac{1}{2}a, \tag{11.5}$

that is, we obtain a double real root. Bearing this in mind as having resolved the case $a^2 - 4b = 0$, we proceed under our hypothesis $a^2 - 4b \ne 0$.

With (6) in place we may write a precise expression for the real parts of the roots of (2) via (3): we have

$a = \Re(a) + i \Im(a), \tag{12}$

and

$\sqrt{a^2 - 4b} = \sqrt{\vert a^2 - 4b \vert} e^{i \arg(a^2 - 4b) / 2}$ $= \sqrt{\vert a^2 - 4b \vert} \cos \dfrac{ \arg (a^2 - 4b)}{2} + i \sqrt{\vert a^2 - 4b \vert} \sin \dfrac{ \arg (a^2 - 4b)}{2}; \tag{13}$

it now follows from (3), (11) that the real parts of $\mu_1$ and $\mu_2$ are

$\Re(\mu_1), \; \Re(\mu_2) = \dfrac{1}{2}\left (-\Re(a) \pm \sqrt{\vert a^2 - 4b \vert} \cos \dfrac{ \arg (a^2 - 4b)}{2} \right ); \tag{14}$

we can in principle, from (14), determine the relationship(s) $a$ and $b$ must satisfy in order that

$\Re(\mu_1), \; \Re(\mu_2) < 0; \tag{15}$

indeed,

$\Re(\mu_1), \Re(\mu_2) = \dfrac{1}{2}\left (-\Re(a) \pm \sqrt{\vert a^2 - 4b \vert} \cos \dfrac{ \arg (a^2 - 4b)}{2} \right ) < 0$ $\iff -\Re(a) + \left \vert \sqrt{\vert a^2 - 4b \vert} \cos \dfrac{ \arg (a^2 - 4b)}{2} \right \vert < 0, \tag{16}$

which logical equivalence ($\Longleftrightarrow$) binds since

$\pm \sqrt{\vert a^2 - 4b \vert} \cos \dfrac{ \arg (a^2 - 4b)}{2} = \left \vert \sqrt{\vert a^2 - 4b \vert} \cos \dfrac{ \arg (a^2 - 4b)}{2} \right \vert \tag{17}$

for at least one choice of $+/-$ in the "$\pm$" sign.

(16) is in fact completely determinative of (15), since it affirms that the largest of the two $\Re(\mu_1)$, $\Re(\mu_2)$ is negative.

A few words on the special case with both $a, b \in \Bbb R$, that is,

$a = \Re(a), b = \Re(b) \in \Bbb R, \tag{18}$

we have

$\arg(a^2 - 4b) = 0, \pi \Longleftrightarrow \dfrac{ \arg (a^2 - 4b)}{2} = 0, \dfrac{\pi}{2} \Longrightarrow \cos \dfrac{ \arg (a^2 - 4b)}{2} = 1, 0; \tag{19}$

with $\cos (\arg(a^2 - 4b) / 2) = 1$ we find that

$ \left \vert \sqrt{\vert a^2 - 4b \vert} \cos \arg \dfrac{(a^2 - 4b)}{2} \right \vert = \sqrt{\vert a^2 - 4b} \vert, \tag{20}$

and so, via (14), (18)

$\mu_1 = \Re(\mu_1), \; \mu_2 = \Re(\mu_2) = \dfrac{1}{2}\left (-a \pm \sqrt{\vert a^2 - 4b \vert} \right ); \tag{21}$

in such a situation we thus see that both roots $\mu_1$ and $\mu_2$ have negative real part when

$a = \vert a \vert > \sqrt{\vert a^2 - 4b \vert}, \tag{22}$

that is, when

$0 < a, \; 0 < b < \dfrac{a^2}{2}; \tag{23}$

on the other hand, with $\cos (\arg(a^2 - 4b) / 2) = 0$, it is even easier to realize that we must have, again with the aid of (14),

$\Re(\mu_1) = \Re(\mu_2) = -\dfrac{1}{2} \Re(a), \tag{24}$

and $\mu_1, \mu_2 = \bar \mu_1$ are a complex conjugate pair; this is consistent with (13), which in the present instance reverts to

$\sqrt{a^2 - 4b} = i \sqrt{\vert a^2 - 4b \vert}, \tag{24}$

and then (11) yields

$\mu_1, \mu_2 = \dfrac{-a \pm i\sqrt{\vert a^2 - 4b \vert}}{2}. \tag{25}$

Answer 2

Let us separate it into two cases :

Case 1 : When $\Re(a^2-4b)+|a^2-4b|=0$, i.e. $\Re(a^2-4b)\le 0$ and $\Im(a^2-4b)=0$, the roots of $z^2+az+b=0$ are given by $$z_1,z_2=\frac{-(\Re(a)+i\Im(a))\pm i\sqrt{4b-a^2}}{2}$$ from which $$\Re(z_1)\lt 0\quad\text{and}\quad \Re(z_2)\lt 0\iff -\frac{\Re(a)}2\lt 0\iff \Re(a)\gt 0$$follows.

Case 2 : When $\Re(a^2-4b)+|a^2-4b|\not=0$, let us use the following lemma (the proof for the lemma is written at the end of the answer) :

Lemma : Let $\Delta :=t^2-4su$. If $s\not=0$ and $\Re(\Delta)+|\Delta|\not=0$, then the roots of $$sx^2+tx+u=0$$ are given by $$x=-\frac{t}{2s}\pm\frac{\Delta+|\Delta|}{2\sqrt 2\ s\sqrt{\Re(\Delta)+|\Delta|}},$$ i.e. $$x=-\frac{t}{2s}\pm \left(\frac{\sqrt{\Re(\Delta)+|\Delta|}}{2\sqrt 2\ s}+i\frac{\Im(\Delta)}{2\sqrt 2\ s\sqrt{\Re(\Delta)+|\Delta|}}\right)$$

From the lemma, the roots of $z^2+az+b=0$ are given by $$\small z_1,z_2=-\frac{\Re(a)+i\Im(a)}{2}\pm\left(\frac{\sqrt{\Re(a^2-4b)+|a^2-4b|}}{2\sqrt 2}+i\frac{\Im(a^2-4b)}{2\sqrt 2\sqrt{\Re(a^2-4b)+|a^2-4b|}}\right)$$ from which $$\begin{align}&\Re(z_1)\lt 0\qquad\text{and}\qquad \Re(z_2)\lt 0 \\\\&\small\iff -\frac{\Re(a)}{2}+\frac{\sqrt{\Re(a^2-4b)+|a^2-4b|}}{2\sqrt 2}\lt 0\quad\text{and}\quad -\frac{\Re(a)}{2}-\frac{\sqrt{\Re(a^2-4b)+|a^2-4b|}}{2\sqrt 2}\lt 0 \\\\&\iff \frac{\Re(a)}{2}\gt \frac{\sqrt{\Re(a^2-4b)+|a^2-4b|}}{2\sqrt 2} \\\\&\iff \sqrt 2\ \Re(a)\gt \sqrt{\Re(a^2-4b)+|a^2-4b|} \\\\&\iff \Re(a)\gt 0\qquad \text{and}\qquad 2(\Re(a))^2\gt \Re(a^2-4b)+|a^2-4b|\end{align}$$ follows.

From the two cases, a necessary and sufficient condition is $$\color{red}{\Re(a)\gt 0\qquad \text{and}\qquad 2(\Re(a))^2\gt \Re(a^2-4b)+|a^2-4b|}$$

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Finally, let us prove the lemma.

Proof for the lemma :

We get $$\begin{align}&s\left(-\frac{t}{2s}\pm\frac{\Delta+|\Delta|}{2\sqrt 2\ s\sqrt{\Re(\Delta)+|\Delta|}}\right)^2+t\left(-\frac{t}{2s}\pm\frac{\Delta+|\Delta|}{2\sqrt 2\ s\sqrt{\Re(\Delta)+|\Delta|}}\right)+u \\\\&=s\left(\frac{t^2}{4s^2}\mp\frac{t(\Delta+|\Delta|)}{2\sqrt 2\ s^2\sqrt{\Re(\Delta)+|\Delta|}}+\frac{(\Delta+|\Delta|)^2}{8s^2(\Re(\Delta)+|\Delta|)}\right) \\\\&\qquad\qquad\qquad\qquad -\frac{t^2}{2s}\pm\frac{t(\Delta+|\Delta|)}{2\sqrt 2\ s\sqrt{\Re(\Delta)+|\Delta|}}+u \\\\&=\frac{t^2}{4s}\mp\frac{t(\Delta+|\Delta|)}{2\sqrt 2\ s\sqrt{\Re(\Delta)+|\Delta|}}+\frac{(\Delta+|\Delta|)^2}{8s(\Re(\Delta)+|\Delta|)}\\\\&\qquad\qquad\qquad\qquad -\frac{t^2}{2s}\pm\frac{t(\Delta+|\Delta|)}{2\sqrt 2\ s\sqrt{\Re(\Delta)+|\Delta|}}+u \\\\&=\frac{-\Delta}{4s}+\frac{(\Delta+|\Delta|)^2}{8s(\Re(\Delta)+|\Delta|)} \\\\&=\frac{-(\Delta+\overline{\Delta})\Delta+\Delta^2+\Delta\overline{\Delta}}{8s(\Re(\Delta)+|\Delta|)} \\\\&=0\qquad\quad\square\end{align}$$