How can we see $ax^2-bx-\bar b x+c,a>0,c \ge 0 $ attains minimum when $x=\frac{\Re b}{a}$ How can we see $ax^2-bx-\bar b x+c, a>0,c\ge 0$ attains minimum (just minimize it over the set of all possible $x$ s.t. the quadratic function is real) when $x=\frac{\Re  b}{a}$ where $\Re b$ is the real part of $b$? I know that $ax^2-bx+c$, where all numbers are real, attains minimum when $x=\frac{b}{2a}$.

You know that $ax^2-bx+c$ attains minimum at $x=\frac{b}{2a}$ (assuming $a,b,c\in\Bbb R$). You can then see that your equation reduces to $ax^2-2\Re bx+c$, which satisfies the assumptions, allowing you to apply the same rule to reach the desired conclusion.
