Complex Conjugates Theorem

Complex Conjugates Theorem

How to use the complex conjugates theorem to find the missing zeros: theorem, example, and its solution.

Theorem

If a + bi is the zero of f(x), then its complex conjugate, a - bi, is also the zero of f(x).

If (a + bi) is the zero of f(x),
then its complex conjugate, (a - bi),
is also the zero of f(x).

Complex conjugates

So both [x - (a + bi)] and [x - (a - bi)]
are the factors of the f(x).

Factor theorem

Example

Find the function that satisfies the given conditions. Highest degree term: x^3, Known zeros: 2, 3 + i

The given zeros are 2 and 3 + i.

So, by the complex conjugate theorem,
the complex conjugate of (3 + i), (3 - i),
is also the zero of the function.

So x = 2, 3 + i, 3 - i.

The highest degree term is x3.
So this function has at most 3 zeros.

So these are the zeros.

x = 2, 3 + i, 3 - i.

The highest degree term is x3.

So, by the factor theorem,
f(x) = (x - 2)[x - (3 + i)][x - (3 - i)].

Factor theorem

To expand [x - (3 + i)][x - (3 - i)],
find the sum and the product of the zeros.

Sum and product of the roots of a quadratic equation

Add the zeros.
Then (3 + i) + (3 - i) = 6.

Multiply the zeros.
Then (3 + i)(3 - i) = 32 + 12.

Complex conjugates

32 = 9
12 = 1

9 + 1 = 10

(3 + i) + (3 - i) = 6
(3 + i)(3 - i) = 10

So [x - (3 + i)][x - (3 - i)]
= x2 - 6x + 10.

Sum and product of the roots of a quadratic equation

So f(x) = (x - 2)(x2 - 6x + 10)

Solve (x - 2)(x2 - 6x + 10).

x(x2 - 6x + 10)
= x3 - 6x2 + 10x

-2(x2 - 6x + 10)
= -2x2 + 12x - 20

Multiplying polynomials

x3 = x3
-6x2 - 2x2 = -8x2
+10x + 12x = +22x
-20 = -20

So f(x) = x3 - 8x2 + 22x - 20.