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Constraints on $x^2 + b x + c = 0$ such that at least one root has a positive real part

Mathematics Asked by kevinkayaks on December 3, 2021

I have a quadratic equation with complex coefficients
$$ 0 = frac{1}{2}x^2 + b x + frac{1}{2}c$$
and would like to determine the conditions on $b$ and $c$ such that at least one of the roots have a positive real part. The roots are
$$ x_pm = -b pm sqrt{b^2-c}. $$

I thought to find real numbers $X$ and $Y$ such that $b^2-c = X +iY$ then to simplify the square root as follows:

$$ sqrt{X + i Y} = sqrt{X^2 + Y^2} Big[ sqrt{frac{1 + frac{X}{sqrt{X^2 + Y^2}}}{2}} + i sqrt{frac{1 -frac{X}{sqrt{X^2 + Y^2}}}{2}}Big], $$ where I assumed $tan^{-1} frac{Y}{X}$ lies in the first quadrant and applied the half angle and Euler formulas.

Then the real part of the roots becomes

$$ text{Re}{x_pm} = -text{Re}{b} pm sqrt{frac{X^2 + Y^2 + X{sqrt{X^2 + Y^2}}}{2}} $$
and the condition that the larger of these is positive is then

$$ X^2 + Y^2 + X{sqrt{X^2 + Y^2}} > 2text{Re}{b}^2 $$

This is not very transparent — very non-linear and complicated. Especially given that I would need to express $X$ and $Y$ in terms of $b$ and $c$ still to constrain these parameters.

Is there some other reasoning I could use to constrain $b$ and $c$ such that at least one root has a positive real part ? Naively I could imagine requiring that $text{Re}{c}$ were negative… but there’s no real obvious reason this should be sufficient considering the mixing terms between complex and real parts in the determinant.

One Answer

Let $b=p+iq$

and $sqrt{b^2-c}=r+is$ where $p,q,r,s$ are real

$implies c=(p+iq)^2-(r+is)^2=cdots$

We need at least one of $-p+r,-p-r$ be $>0$

Answered by lab bhattacharjee on December 3, 2021

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