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 $x=\frac{b}{2a}$.
 A: Call $B=b/a$ and $C=c/a$; the problem is the same as determining the point of minimum of $x^2-Bx-\bar{B}x+C$.
First of all, try and determine, for $x\in\mathbb{C}$ when $x^2-Bx-\bar{B}x+C$ is real (and also the original function will be real); since
$$
\overline{x^2-Bx-\bar{B}x+C}=\bar{x}^2-\bar{B}\bar{x}-B\bar{x}+C
$$
you need
$$
x^2-Bx-\bar{B}x=\bar{x}^2-\bar{B}\bar{x}-B\bar{x}
$$
that can be rewritten as
$$
(x^2-\bar{x}^2)-B(x-\bar{x})-\bar{B}(x-\bar{x})=0
$$
or
$$
(x-\bar{x})((x+\bar{x})-B-\bar{B})=0
$$
Clearly, this happens when either $x=\bar{x}$ (that is, $x$ is real) or
$$
x+\bar{x}=B+\bar{B}
$$
that is, $\Re(x)=\Re(B)$.
The minimum for real $x$ is attained at $(B+\bar{B})/2=\Re(B)$ and the value is
$$
4C-(B+\bar{B})^2
$$
Setting for simplicity $B=\beta+i\gamma$, we have, for $x=\beta+iy$,
$$
\beta^2-y^2+2i\beta y-2\beta(\beta+iy)+C=C-\beta^2-y^2
$$
and it is clear that we have no minimum at all, because $y$ can be arbitrary.
A: The easiest thing is to think about your function of $x$ as a function of $u=\Re x$ and $v=\Im x$, the real and imaginary parts of $x$. Just take the partial derivatives with respect to these variables, equate them to 0, and solve the system 
A: 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.
A: 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.
