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 $x=\frac{b}{2a}$.

Let $\beta = \text{Real}(b)$. Notice
$$
\begin{aligned}
ax^2 - bx - \bar{b}x + c &= ax^2 - (b + \bar{b}) x + c \\
&= ax^2 - 2 \beta x + c \\
&= a\left( x^2 - 2 \frac{\beta}{a} x\right) + c \\
&= a \left( x - \frac{\beta}{a} \right)^2 + c - \frac{\beta^2}{a}
\end{aligned}
$$
where the last equality follows by completing the square. With $a > 0$ and $a, \beta, c \in \Bbb{R}$, notice that the two things you want to "control" about the original equation are "controlled" by the term $\left( x - \frac{\beta}{a}\right)^2$; that is, the expression is real if and only if $\left( x - \frac{\beta}{a}\right)^2$ is real, and the expression is minimized if and only if $\left( x - \frac{\beta}{a}\right)^2$ is minimized. If you can show that $\left( x - \frac{\beta}{a}\right)^2$ is real if and only if $x$ is real, then the minimization part follows almost immediately.
