Necessary and sufficient conditions for a quadratic polynomial with complex coefficients to have both roots with negative real parts 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
$$
B_1 > 0, \qquad B_1^2 C_1 + B_1 B_2 C_2 - A C_2^2 > 0.
$$
 A: 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
$$\color{red}{\Re(a)\gt 0\qquad \text{and}\qquad 2(\Re(a))^2\gt \Re(a^2-4b)+|a^2-4b|}$$

Finally, let us prove the lemma.
Proof for the lemma : 
We get
$$\begin{align}&s\left(-\frac{t}{2s}\pm\frac{\Delta+|\Delta|}{2\sqrt 2\ s\sqrt{\Re(\Delta)+|\Delta|}}\right)^2+t\left(-\frac{t}{2s}\pm\frac{\Delta+|\Delta|}{2\sqrt 2\ s\sqrt{\Re(\Delta)+|\Delta|}}\right)+u
\\\\&=s\left(\frac{t^2}{4s^2}\mp\frac{t(\Delta+|\Delta|)}{2\sqrt 2\ s^2\sqrt{\Re(\Delta)+|\Delta|}}+\frac{(\Delta+|\Delta|)^2}{8s^2(\Re(\Delta)+|\Delta|)}\right)
\\\\&\qquad\qquad\qquad\qquad -\frac{t^2}{2s}\pm\frac{t(\Delta+|\Delta|)}{2\sqrt 2\ s\sqrt{\Re(\Delta)+|\Delta|}}+u
\\\\&=\frac{t^2}{4s}\mp\frac{t(\Delta+|\Delta|)}{2\sqrt 2\ s\sqrt{\Re(\Delta)+|\Delta|}}+\frac{(\Delta+|\Delta|)^2}{8s(\Re(\Delta)+|\Delta|)}\\\\&\qquad\qquad\qquad\qquad -\frac{t^2}{2s}\pm\frac{t(\Delta+|\Delta|)}{2\sqrt 2\ s\sqrt{\Re(\Delta)+|\Delta|}}+u
\\\\&=\frac{-\Delta}{4s}+\frac{(\Delta+|\Delta|)^2}{8s(\Re(\Delta)+|\Delta|)}
\\\\&=\frac{-(\Delta+\overline{\Delta})\Delta+\Delta^2+\Delta\overline{\Delta}}{8s(\Re(\Delta)+|\Delta|)}
\\\\&=0\qquad\quad\square\end{align}$$
