Complex roots of a quadratic equation

Apr 2010
487
9
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
Aug 2008
12,897
5,001
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.
 
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Apr 2010
487
9
Ahh, thanks. So would my solution be acceptable, or should I simplify it like yours?
 

Prove It

MHF Helper
Aug 2008
12,897
5,001
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...
 
Apr 2010
487
9
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
Aug 2006
22,507
8,664
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
Apr 2005
20,249
7,909
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!
 
Dec 2009
1,506
434
Russia
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...
 
Apr 2010
487
9
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
May 2006
12,028
6,341
Lexington, MA (USA)
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 \(\displaystyle z \:=\:p + qi\) 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: .\(\displaystyle \overline z \:=\:p - qi\)

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

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


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

 
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