# Complex roots of a quadratic equation

#### 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 $$\displaystyle ax^2 + bx + c = 0$$ with a, b, c being real coefficients, then so is the conjugate of 'z'.

My attempt:

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

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

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

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

MHF Helper
The roots of $$\displaystyle ax^2 + bx + c = 0$$ are

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

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

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

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

So the two roots are complex conjugates.

• Glitch

#### Glitch

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

#### Prove It

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

#### Glitch

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

MHF Helper
The question:
Use the properties of the complex conjugate to show that if the complex number 'z' is a root of a quadratic equation $$\displaystyle ax^2 + bx + c = 0$$ 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 $$\displaystyle a$$ is a real number then $$\displaystyle \overline{a}=a$$.
The conjugate of a sum is the sum of conjugates.
Therefore we have:

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

#### HallsofIvy

MHF Helper
The question:

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

My attempt:

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

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

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

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

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

#### Glitch

Fair point. So if I add:

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

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

#### Soroban

MHF Hall of Honor
Hello, Glitch!

I have a very primitive solution . . .

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

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

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

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

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

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

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

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

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

• Glitch