Complex Conjugates Theorem
How to use the complex conjugates theorem to find the roots of an equation: theorem, 1 example, and its solution.
Theorem
Theorem
If [a + bi] is the zero of f(x),
then its conjugate [a - bi]
is also the zero of f(x).
This is the complex conjugates theorem.
Example
Example
Solution
The highest degree term is x3.
Then there are 3 zeros.
The known zeros are
x = 2 and 3 + i.
Then, by the complex conjugates theorem,
the conjugate of 3 + i, 3 - i,
is also the zero.
So x = 2, 3 + i, 3 - i
are the zeros.
The zeros are
x = 2, 3 + i, 3 - i.
The highest degree term is x3.
The coefficient is 1.
So f(x) = (x - 2)(x - [3 + i])(x - [3 - i]).
Factor Theorem
Expand (x - [3 + i])(x - [3 - i]).
Instaed of directly expanding this,
use the sum and the product of the roots.
(3 + i) + (3 - i) = 6
The product of the roots is
(3 + i)(3 - i).
(3 + i)(3 - i) = 32 + 12
Formula: Divide Complex Numbers
32 + 12 = 9 + 1
9 + 1 = 10
Write the front factor (x - 2).
(3 + i) + (3 - i) = 6
(3 + i)(3 - i) = 10
Then
(x - [3 + i])(x - [3 - i])
= (x2 - 6x + 10).
So
(x - 2)(x - [3 + i])(x - [3 - i])
= (x - 2)(x2 - 6x + 10).
-2⋅x2 = -2x2
-2⋅(-6x) = +12x
-2⋅(+10) = -20
So
(x - 2)(x2 - 6x + 10)
= x3 - 6x2 + 10x - 2x2 + 12x - 20.
-6x2 - 2x2 = -8x2
+10x + 12x = +22x
So
x3 - 8x2 + 22x - 20
is the answer.