Complex roots of a quadratic equation

**Glitch** · July 30th 2010, 04:05 AM

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!

**Prove It** · July 30th 2010, 04:09 AM

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.

**Glitch** · July 30th 2010, 04:18 AM

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

**Prove It** · July 30th 2010, 04:22 AM

Originally Posted by Glitch

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

To be honest, I can't really follow your solution...

**Glitch** · July 30th 2010, 04:27 AM

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.

**Plato** · July 30th 2010, 04:39 AM

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:

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

**HallsofIvy** · July 30th 2010, 04:58 AM

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!

**Also sprach Zarathustra** · July 30th 2010, 05:04 AM

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...

**Glitch** · July 30th 2010, 05:04 AM

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?

**Soroban** · July 30th 2010, 09:44 AM

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

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

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

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