In mathematics, the complex conjugate root theorem is one of the important theorems to find the complex roots of a polynomial.

This theorem points out that complex roots of polynomial equations with real coefficients occur in complex conjugate pairs.

We use various methods to find the zeros of polynomials such as factorization method, synthetic division etc.

Complex conjugate theorem tells that if a polynomial have complex zero then conjugate of that zero is also its one of the root. **For Example:** if 3 + 4i is a zero of quadratic polynomial P(x) then 3 - 4i is another zero of P(x). It is used to find the complex roots.

In complex plane, complex numbers can be represented as vectors or as points when we are dealing with graphs of complex roots.

In this section we will study about complex conjugate theorem with proof and some solved examples.

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Th solution of any equation can be find by using fundamental theorem of algebra and the conjugate root theorem.

According fundamental theorem, if a polynomial is of degree, m greater than equal to one with complex coefficients then there exist m solutions of the equation.

And according to the conjugate root theorem, if a polynomial g(x) of degree m greater than equal to two with real coefficients, then if c + id where d is not equal to zero, is a zero of the equation g(x) = 0, then c - id is a zero also.

According to complex conjugate theorem, a polynomial F(x) with real coefficients, and c + id is a root of F(x) where a and b are real numbers, then complex conjugate of c - id is also one of its root.

We can also find the polynomial equation if roots are given by using factorization method. If z = c is a complex root of f(z) = 0 then (z - c) and (z - $\bar{c}$) must be factors of f(z).

Suppose we have to find polynomial equation has real coefficients and complex root 1 - i.

According to complex conjugate theorem, complex conjugate of 1 - i i.e. 1 + i also solution of polynomial.

Using linear factorization: (x - (1 - i)) and (x - (1 + i)) are factors of polynomial. Let f(x) be the required polynomial, then:

$f(x)$ = $(x - (1 - i)) (x - (1 + i))$ = $((x - 1) + i) ((x - 1) - i)$

$f(x)$ = $((x - 1)^2 - i^2)$

$f(x)$ = (x$^2$ + 1 - 2x + 1)

$f(x)$ = x$^2$ - 2$x$ + 2. Consider a polynomial f(x) = $b_0 + b_1x + b_2x^2 + .....+ b_nx^n$

where $b_0, b_1, b_2,....., b_n$ are real.

Suppose any complex number say Z is a root of f(x), that is f(z) = 0.

= $b_0 + b_1z + b_2z^2 + .....+ b_nz^n$ = 0

Since all the given coefficients are real, we have

$\overline{b_0 + b_1z + b_2z^2 + .....+ b_nz^n}$ = $b_0 + b_1 \overline{z} + b_2 \overline{z^2} + ..... + b_n \overline{z^n}$

It follows that

$b_0 + b_1 \bar{z} + b_2 \bar{z^2} + .....+ b_n \bar{z^n}$ = 0

Thus for any complex root z, its complex conjugate $\bar{z}$ is also a root. Hence proved.

Let us prove with the help of an example. Find the roots of a cubic polynomial f(x) = x$^3$ + 7x - 3x$^2$ - 5.

**Solution:** $f(x)$ = x$^3$ + 7x - 3x$^2$ - 5

Degree of polynomial = 3 (odd). It means it must have at least one real root.

Apply Trial and Error method to find the real root.

Put x = 1, f(1) = 1$^3$ + 7 . 1 - 3. 1$^2$ - 5 = 1 + 7 - 3 - 5 = 0

This shows that x = 1 is one of the zero of f(x)

To find other zeros divide $f(x)$ by x - 1, we obtain

$f(x)$ = (x - 1)(x$^2$ - 2x + 5)

x$^2$ - 2x + 5 = (x - (1 $\pm$ 2i) (Use quadratic formula)

Therefore the zeros of f(x) are 1, 1 - 2i and 1 + 2i.

**Some of the problems on complex conjugate theorem are given below:**

**Example 1:** Create a polynomial that has zeros 2 and 2 + i.

**Solution:** According to complex conjugate theorem, conjugate of 2 + i is also its root i.e. 2 - i. So required polynomial having 3 roots, 2, 2 + i and 2 - i.

Let f(x) be any polynomial with roots 2, 2 + i and 2 - i.

So $f(x)$ = (x - 2) (x - (2 + i)) (x - (2 - i))

$f(x)$ = (x - 2) ((x - 2) - i)) ((x - 2) + i))

$f(x)$ = (x - 2) ((x - 2)$^2$ - i$^2$)) (using identity (a - b)(a + b) = a$^2$ - b$^2$ and $i^2$ = -1)

$f(x)$ = (x - 2) (x$^2$ + 2$^2$ - 4x - (-1))

$f(x)$ = (x - 2) (x$^2$ + 4 - 4x + 1)

$f(x)$ = x$^3$ + 4x - 4x$^2$ + x -2x$^2$ - 8 + 8x - 2

$f(x)$ = x$^3$ - 6x$^2$ + 13x - 10

**Example 2:** Find a second degree polynomial equation that has root of 3 + i.

**Solution:**

__Step 1:__ Determine the roots

Since given root is complex, so apply conjugate theorem to determine the second root.

Complex conjugate of 3 + i is 3 - i.

Because 3 + i is a root, its complex conjugate 3 - i is also a root.

The solution set of polynomial is {3 - i, 3 + i}

__Step 2:__ Write the linear factorization of the polynomial and form the equation.

Since 3 + i and 3 - i are roots, (x - (3 + i)) and (x - (3 - i)) are factors.

The polynomial equation is the product of factors i.e.

(x - (3 + i))(x - (3 - i)) = ((x - 3) - i) ((x - 3) + i)

= $((x - 3)^2 - i^2)$

= (x$^2$ + 9 - 6x + 1)

= x$^2$ - 6x + 10

The equation x$^2$ - 6x + 10 will have root of 3 + i and 3 - i.

According fundamental theorem, if a polynomial is of degree, m greater than equal to one with complex coefficients then there exist m solutions of the equation.

And according to the conjugate root theorem, if a polynomial g(x) of degree m greater than equal to two with real coefficients, then if c + id where d is not equal to zero, is a zero of the equation g(x) = 0, then c - id is a zero also.

According to complex conjugate theorem, a polynomial F(x) with real coefficients, and c + id is a root of F(x) where a and b are real numbers, then complex conjugate of c - id is also one of its root.

We can also find the polynomial equation if roots are given by using factorization method. If z = c is a complex root of f(z) = 0 then (z - c) and (z - $\bar{c}$) must be factors of f(z).

Suppose we have to find polynomial equation has real coefficients and complex root 1 - i.

According to complex conjugate theorem, complex conjugate of 1 - i i.e. 1 + i also solution of polynomial.

Using linear factorization: (x - (1 - i)) and (x - (1 + i)) are factors of polynomial. Let f(x) be the required polynomial, then:

$f(x)$ = $(x - (1 - i)) (x - (1 + i))$ = $((x - 1) + i) ((x - 1) - i)$

$f(x)$ = $((x - 1)^2 - i^2)$

$f(x)$ = (x$^2$ + 1 - 2x + 1)

$f(x)$ = x$^2$ - 2$x$ + 2. Consider a polynomial f(x) = $b_0 + b_1x + b_2x^2 + .....+ b_nx^n$

where $b_0, b_1, b_2,....., b_n$ are real.

Suppose any complex number say Z is a root of f(x), that is f(z) = 0.

= $b_0 + b_1z + b_2z^2 + .....+ b_nz^n$ = 0

Since all the given coefficients are real, we have

$\overline{b_0 + b_1z + b_2z^2 + .....+ b_nz^n}$ = $b_0 + b_1 \overline{z} + b_2 \overline{z^2} + ..... + b_n \overline{z^n}$

It follows that

$b_0 + b_1 \bar{z} + b_2 \bar{z^2} + .....+ b_n \bar{z^n}$ = 0

Thus for any complex root z, its complex conjugate $\bar{z}$ is also a root. Hence proved.

Let us prove with the help of an example. Find the roots of a cubic polynomial f(x) = x$^3$ + 7x - 3x$^2$ - 5.

Degree of polynomial = 3 (odd). It means it must have at least one real root.

Apply Trial and Error method to find the real root.

Put x = 1, f(1) = 1$^3$ + 7 . 1 - 3. 1$^2$ - 5 = 1 + 7 - 3 - 5 = 0

This shows that x = 1 is one of the zero of f(x)

To find other zeros divide $f(x)$ by x - 1, we obtain

$f(x)$ = (x - 1)(x$^2$ - 2x + 5)

x$^2$ - 2x + 5 = (x - (1 $\pm$ 2i) (Use quadratic formula)

Therefore the zeros of f(x) are 1, 1 - 2i and 1 + 2i.

Let f(x) be any polynomial with roots 2, 2 + i and 2 - i.

So $f(x)$ = (x - 2) (x - (2 + i)) (x - (2 - i))

$f(x)$ = (x - 2) ((x - 2) - i)) ((x - 2) + i))

$f(x)$ = (x - 2) ((x - 2)$^2$ - i$^2$)) (using identity (a - b)(a + b) = a$^2$ - b$^2$ and $i^2$ = -1)

$f(x)$ = (x - 2) (x$^2$ + 2$^2$ - 4x - (-1))

$f(x)$ = (x - 2) (x$^2$ + 4 - 4x + 1)

$f(x)$ = x$^3$ + 4x - 4x$^2$ + x -2x$^2$ - 8 + 8x - 2

$f(x)$ = x$^3$ - 6x$^2$ + 13x - 10

Since given root is complex, so apply conjugate theorem to determine the second root.

Complex conjugate of 3 + i is 3 - i.

Because 3 + i is a root, its complex conjugate 3 - i is also a root.

The solution set of polynomial is {3 - i, 3 + i}

Since 3 + i and 3 - i are roots, (x - (3 + i)) and (x - (3 - i)) are factors.

The polynomial equation is the product of factors i.e.

(x - (3 + i))(x - (3 - i)) = ((x - 3) - i) ((x - 3) + i)

= $((x - 3)^2 - i^2)$

= (x$^2$ + 9 - 6x + 1)

= x$^2$ - 6x + 10

The equation x$^2$ - 6x + 10 will have root of 3 + i and 3 - i.

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