1. ## Complex roots of a quadratic equation

The question:

Use the properties of the complex conjugate to show that if the complex number 'z' is a root of a quadratic equation $ax^2 + bx + c = 0$ with a, b, c being real coefficients, then so is the conjugate of 'z'.

My attempt:

Let z = $z_{1} \pm z_2i$

$\frac{-b \pm \sqrt{b^2 - 4ac}}{2a} = z_{1} \pm z_2i$

$\frac{1}{2a}(-b \pm \sqrt{b^2 - 4ac}) = z_{1} \pm z_2i$

Basically, I changed the quadratic formula to look like a complex number in Cartesian form. Is this an acceptable answer? My textbook doesn't have solutions for 'show' and 'prove' questions.

Thanks!

2. The roots of $ax^2 + bx + c = 0$ are

$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$.

If $b^2 - 4ac < 0$ then that means $\sqrt{b^2 - 4ac}$ is imaginary. So we can say $\sqrt{b^2 - 4ac} = ni$, where $n$ is some multiple of $i$.

So $x = \frac{-b \pm n i}{2a}$

$= -\frac{b}{2a} \pm \left(\frac{n}{2a}\right)i$.

So the two roots are complex conjugates.

3. Ahh, thanks. So would my solution be acceptable, or should I simplify it like yours?

4. Originally Posted by Glitch
Ahh, thanks. So would my solution be acceptable, or should I simplify it like yours?

5. It's pretty much the same as yours, except you replaced the square root with 'ni'. I was trying to show that the general form of a complex number is equivalent to a simplified quadratic formula.

6. Originally Posted by Glitch
The question:
Use the properties of the complex conjugate to show that if the complex number 'z' is a root of a quadratic equation $ax^2 + bx + c = 0$ with a, b, c being real coefficients, then so is the conjugate of 'z'.
The standard way of proving the is to use properties of the complex conjugate.
If $a$ is a real number then $\overline{a}=a$.
The conjugate of a sum is the sum of conjugates.
Therefore we have:

$\overline {az^2 + bz + c} = \overline 0 = 0\, \Rightarrow \,a\overline z ^2 + b\overline z + c = 0$

7. Originally Posted by Glitch
The question:

Use the properties of the complex conjugate to show that if the complex number 'z' is a root of a quadratic equation $ax^2 + bx + c = 0$ with a, b, c being real coefficients, then so is the conjugate of 'z'.

My attempt:

Let z = $z_{1} \pm z_2i$

$\frac{-b \pm \sqrt{b^2 - 4ac}}{2a} = z_{1} \pm z_2i$

$\frac{1}{2a}(-b \pm \sqrt{b^2 - 4ac}) = z_{1} \pm z_2i$
But at this point you haven't yet said that $\sqrt{b^2- 4ac}$ is an imaginary number!

Basically, I changed the quadratic formula to look like a complex number in Cartesian form. Is this an acceptable answer? My textbook doesn't have solutions for 'show' and 'prove' questions.

Thanks!

8. Originally Posted by Plato
The standard way of proving the is to use properties of the complex conjugate.
If $a$ is a real number then $\overline{a}=a$.
The conjugate of a sum is the sum of conjugates.
Therefore we have:

$\overline {az^2 + bz + c} = \overline 0 = 0\, \Rightarrow \,a\overline z ^2 + b\overline z + c = 0$
...and it is true for every polynomial...

9. Fair point. So if I add:

When $b^2 - 4ac < 0$, $\sqrt{b^2 - 4ac} = ni, n \in R$

And replace the square root with the imaginary part 'n', would that suffice?

10. Hello, Glitch!

I have a very primitive solution . . .

Use the properties of the complex conjugate to show that
if the complex number $z$ is a root of a quadratic equation
$f(x) \:=\:ax^2 + bx + c = 0$ with $a, b, c$ real coefficients,
then so is the conjugate of $z.$

We are told that $z \:=\:p + qi$ is a solution of the quadratic.

. . Hence: . $a(p+qi)^2 + b(p+qi) + c \:=\:0$

This simplifies to: . $(ap^2 - aq^2 + bp + c) + q(2ap + b)i \;=\;0$

. . And we have: . $\begin{Bmatrix}ap^2 - aq^2 + bp + c \;=\;0 \\ q(2ap + b) \;=\;0 \end{Bmatrix}$

The conjugate of $z$ is: . $\overline z \:=\:p - qi$

Consider $f(\overline z) \;=\;a(p-qi)^2 + b(p-qi) + c$

. . . . . . . $f(\overline z) \;=\;\underbrace{(ap^2 - aq^2 + bp + c)}_{\text{This is 0}} - \underbrace{q(2ap + b)}_{\text{This is 0}}i$

$\text{Therefore: }\:f(\overline z) \,=\,0\:\text{ and }\:\overline z\text{ is also a solution.}$