What is a complex constant and how do I use it?

Question

I have a question I am trying to understand: "Let $b$ and $c$ be complex constants such that $z^2+bz+c=0$ has two different real roots. Show that $b$ and $c$ are real."

My biggest problem here is that I don't know what a complex constant is; I have tried to google it, however all I am getting for results is computer programming information, which I am finding unhelpful. Could someone please explain to me what exactly this is and how I can use it to solve the above problem?

Answer 1

Using the Hint from Semiclassical with the two real roots $r_1 \neq r_2$: \begin{align} r_1^2+r_1(b_1 +ib_2) + (c_1 + ic_2) &= 0\\ r_2^2+r_2(b_1 +ib_2) + (c_1 + ic_2) &= 0 \end{align}

Summing up the imaginary parts

\begin{align} i(r_1b_2 + c_2) &= 0\\ i(r_2b_2 + c_2) &= 0 \end{align}

So you get either $b_2 = c_2 =0$ or $r_1 = r_2$. Last one would be a contradiction to the statement above. Therefore $b_2 = c_2 =0$ and $b$ and $c$ are real.

Answer 2

Hint 1: Suppose you have a polynomial with two distinct real roots be labelled as $r_1,r_2$. How would you write this polynomial, and how does this compare with the form given?

Hint 2: Does multiplying or adding real numbers ever produce a complex number?

Answer 3

$z=\dfrac{-b\pm\sqrt\Delta}{2}$, $\Delta=b^2-4c$. In order to have two real solutions you must have $\Delta\gt0$ and $b$ real. $c\in\mathbb{R}$, follows.

Answer 4

As already mentioned by others, in this context complex constants simply means complex numbers.

Hint to solve the problem: Try to use Vieta's formulas.

$b=-z_1-z_2$
$c=z_1z_2$
where $z_{1,2}$ are the roots