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
$$ X^2 + Y^2 + X{\sqrt{X^2 + Y^2}} > 2\text{Re}\{b\}^2 $$
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.
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 $>0$