# sign of roots of a quadratic equation with complex coefficients.

Consider $x^2+ax+b=0$, where $x$ is the variable and $a,b$ are complex coefficients. Is there any condition on $a$ and $b$ which makes sure the roots of the equation have negative real parts?

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A necessary condition is $\Re a>0$, a sufficient condition is $4|b|<(\Re a)^2$. – Hagen von Eitzen Jan 10 '13 at 13:48
@HagenvonEitzen: How did you obtain this inequalities? Here I have a similler question. Can you help me to solve it? Thank you. math.stackexchange.com/questions/985518/… – Nilan Oct 22 '14 at 7:44
@Nilan $-a$ is the sum of the two roots, hence $\Re a>0$ is necessary. – Hagen von Eitzen Oct 22 '14 at 10:30
@HagenvonEitzen: I got that. How did you get the sufficient condition? – Nilan Oct 23 '14 at 7:46

Let $\,z_1:=a+bi\,,\,z_2:=c+di\in\Bbb C\,$ be the two roots, then using Viete's formulae:
$$ac-bd+(ad+bc)i=z_1z_2=b:=x+yi$$
$$a+c+(b+d)i=z_1+z_2=-a:=-p-qi$$