AOS3 Topic 5: Complex Polynomials

Definition:

A complex polynomial is an algebraic expression of the form:

$$ P(z) = a_n z^n + a_{n-1} z^{n-1} + \ldots + a_1 z + a_0 $$

where \( z \) is a complex variable, \( a_0, a_1, \ldots, a_n \) are complex coefficients, ehere \(a_n \neq 0\) and \( n \) is a non-negative integer.

Degree:

The degree of a complex polynomial is the highest power of the variable \( z \) in the polynomial. For example, in the polynomial \( P(z) \), the degree is \( n \).

Roots:

The roots of a complex polynomial are the values of \( z \) for which the polynomial evaluates to zero. These roots can be real or complex numbers. The Fundamental Theorem of Algebra states that every non-constant complex polynomial has at least one complex root.


Created with GeoGebra®, Steve Phelps, Link




Created with GeoGebra®, Steve Phelps, Link


When we divide the polynomial \(P(z)\) by the polynomial \(D(z)\) we obtain two polynomials, \(Q(z)\) the quotient and \(R(z)\) the remainder, such that

\(P(z) = D(z)Q(z) + R(z)\) and either \(R(z) = 0\) or \(R(z)\) has degree less than \(D(z)\).

If \(R(z) = 0\), then \(D(z)\) is a factor of \(P(z)\).

The remainder theorem and the factor theorem are true for polynomials over C.

  • Remainder theorem: Let \( \alpha \in \mathbb{C} \). When a polynomial \(P(z)\) is divided by \(z − α\), the remainder is \(P(α)\).
  • Factor theorem: Let \( \alpha \in \mathbb{C} \). Then \(z − α\) is a factor of a polynomial \(P(z)\) if and only if \(P(α) = 0\).

Conjugate root theorem:

Let \(P(z)\) be a polynomial with real coefficients. If \(a + bi\) is a solution of the equation \(P(z) = 0\), with \(a\) and \(b\) real numbers, then the complex conjugate \(a − bi\) is also a solution.

Proof: We will prove the theorem for quadratics, as it gives the idea of the general proof. Let \( P(z) = az^2 + bz + c \), where \( a, b, c \in \mathbb{R} \) and \( a \neq 0 \). Assume that \( \alpha \) is a solution of the equation \( P(z) = 0 \). Then \( P(\alpha) = 0 \). That is, \( a\alpha^2 + b\alpha + c = 0 \). Take the conjugate of both sides of this equation and use properties of conjugates: \( a\alpha^2 + b\alpha + c = 0 \). \( a\alpha^2 + b\alpha + c = 0 \). \( a(\alpha^2) + b\alpha + c = 0 \) since \( a, b \), and \( c \) are real numbers. \( a(\alpha)^2 + b\alpha + c = 0 \). Hence \( P(\alpha) = 0 \). That is, \( \alpha \) is a solution of the equation \( P(z) = 0 \). If a polynomial \( P(z) \) has real coefficients, then using this theorem we can say that the complex solutions of the equation \( P(z) = 0 \) occur in conjugate pairs.

Factorization of cubic polynomials

Factorization of cubic polynomials involves breaking down a cubic polynomial equation into its linear factors. Over the complex numbers, every cubic polynomial can be written as the product of three linear factors. This means that if we have a cubic polynomial \( P(z) \), we can express it as:

\[ P(z) = (z - \alpha)(z - \beta)(z - \gamma) \]

where \( \alpha, \beta, \gamma \) are the roots of the polynomial equation \( P(z) = 0 \).

If the coefficients of the cubic polynomial are real, then according to the conjugate root theorem, complex roots occur in conjugate pairs. Therefore, if the coefficients are real, at least one of the linear factors must be real. This implies that even if the roots are complex, their conjugates must also be present as factors, ensuring that the resulting polynomial has real coefficients.

The fundamental theorem of algebra

The fundamental theorem of algebra, often attributed to Gauss (1799), is a crucial result in the field of mathematics, particularly in the study of polynomials and complex numbers.

Fundamental Theorem of Algebra: Every polynomial \( P(z) = a_nz^{n} + a_{n-1}z^{n-1} + \ldots + a_1z + a_0 \) of degree \( n \), where \( n \geq 1 \) and the coefficients \( a_i \) are complex numbers, has at least one linear factor in the complex number system.

This theorem implies that given any polynomial \( P(z) \) of degree \( n \geq 1 \), we can factorize \( P(z) \) as:

\[ P(z) = (z - \alpha_1)Q(z) \]

for some \( \alpha_1 \in \mathbb{C} \) and some polynomial \( Q(z) \) of degree \( n - 1 \).

Linear factor in \(\mathbb{C}\)

A polynomial of degree \( n \) can be factorized into \( n \) linear factors in \( \mathbb{C} \), i.e.,

\[ P(z) = a_n(z - \alpha_1)(z - \alpha_2)(z - \alpha_3) \ldots (z - \alpha_n) \]

where \( \alpha_1, \alpha_2, \alpha_3, \ldots, \alpha_n \in \mathbb{C} \).

A polynomial equation can be solved by first rearranging it into the form \( P(z) = 0 \), where \( P(z) \) is a polynomial, and then factorizing \( P(z) \) and extracting a solution from each factor.

If \( P(z) = (z - \alpha_1)(z - \alpha_2) \ldots (z - \alpha_n) \), then the solutions of \( P(z) = 0 \) are \( \alpha_1, \alpha_2, \ldots, \alpha_n \).

The solutions of the equation \( P(z) = 0 \) are also referred to as the zeros or the roots of the polynomial \( P(z) \).

Note: A polynomial equation may have repeated solutions. For example, the equation \( (z - 2)^3 = 0 \) has a repeated solution \( z = 2 \). This solution has a multiplicity of 3.

Example 1

Using factor theorem

Factorise \( P(z) = z^3 + z^2 + 4 \).

Solution:

Use the factor theorem to find the first factor:

  • P(-1) = -1 + 1 + 4 = 0
  • P(-2) = -8 + 4 + 4 = 0

Therefore \( z + 2 \) is a factor. We obtain \( P(z) = (z + 2)(z^2 - z + 2) \) by division.

We can factorise \( z^2 - z + 2 \) by completing the square: \[ z^2 - z + 2 = \left(z^2 - z + \frac{1}{4}\right) + 2 - \frac{1}{4} = \left(z - \frac{1}{2}\right)^2 - \frac{7}{4}i^2 = \left(z - \frac{1}{2} + \frac{\sqrt{7}}{2}i\right)\left(z - \frac{1}{2} - \frac{\sqrt{7}}{2}i\right) \]

Hence \( P(z) = (z + 2)\left(z - \frac{1}{2} + \frac{\sqrt{7}}{2}i\right)\left(z - \frac{1}{2} - \frac{\sqrt{7}}{2}i\right) \).

Example 2

Let \( P(z) = z^3 - 3z^2 + 5z - 3 \).

a) Use the factor theorem to show that \( z - 1 + \sqrt{2}i \) is a factor of \( P(z) \):

Find the other linear factors of \(P(z)\).

Solution

a) To show that \( z - (1 - \sqrt{2}i) \) is a factor, we must check that \( P(1 - \sqrt{2}i) = 0 \).

We have: \[ P(1 - \sqrt{2}i) = (1 - \sqrt{2}i)^3 - 3(1 - \sqrt{2}i)^2 + 5(1 - \sqrt{2}i) - 3 \]

Therefore, \( z - (1 - \sqrt{2}i) \) is a factor of \( P(z) \).

b) Since the coefficients of \( P(z) \) are real, the complex linear factors occur in conjugate pairs, so \( z - (1 + \sqrt{2}i) \) is also a factor.

To find the third linear factor, first multiply the two complex factors together: \[ (z - (1 - \sqrt{2}i))(z - (1 + \sqrt{2}i)) = z^2 - (1 - \sqrt{2}i)z - (1 + \sqrt{2}i)z + (1 - \sqrt{2}i)(1 + \sqrt{2}i) \] \[ = z^2 - 2z + 3 \]

Therefore, by inspection, the linear factors of \( P(z) = z^3 - 3z^2 + 5z - 3 \) are \( z - (1 - \sqrt{2}i) \), \( z - (1 + \sqrt{2}i) \) and \( z - 1 \).

Example 3

Factorization of higher degree polynomials

Polynomials of the form \( z^6 - a^6 \) are considered in the following example:

Example:

Factorize \(P(z)=z^{6}-1\).

Solution:

Given \( P(z) = z^6 - 1 = (z^3 + 1)(z^3 - 1) \)

We next factorize \( z^3 + 1 \) and \( z^3 - 1 \)

We have:

\( z^3 + 1 = (z + 1)(z^2 - z + 1) \)

\( = (z + 1)\left(z^2 - z + \frac{1}{4}\right) + 1 - \frac{1}{4} \)

\( = (z + 1)\left(z - \frac{1}{2}\right)^2 - \frac{3}{4}i^2 \)

\( = (z + 1)\left(z - \frac{1}{2} + \frac{\sqrt{3}}{2}i\right)\left(z - \frac{1}{2} - \frac{\sqrt{3}}{2}i\right) \)

By a similar method, we have:

\( z^3 - 1 = (z - 1)(z^2 + z + 1) \)

\( = (z - 1)\left(z + \frac{1}{2} + \frac{\sqrt{3}}{2}i\right)\left(z + \frac{1}{2} - \frac{\sqrt{3}}{2}i\right) \)

Therefore,

\( z^6 - 1 = (z + 1)(z - 1)\left(z - \frac{1}{2} + \frac{\sqrt{3}}{2}i\right)\left(z - \frac{1}{2} - \frac{\sqrt{3}}{2}i\right)\left(z + \frac{1}{2} + \frac{\sqrt{3}}{2}i\right)\left(z + \frac{1}{2} - \frac{\sqrt{3}}{2}i\right) \)

Example 4

Solving complex equations: Fundamental theorem of algebra

Solve the equation over \(\mathbb{C} \) :\( z^3 + 3z^2 + 7z + 5 = 0 \)

Solution;

Let \( P(z) = z^3 + 3z^2 + 7z + 5 \).

Then \( P(-1) = 0 \), so \( z + 1 \) is a factor, by the factor theorem.

\( P(z) = (z + 1)(z^2 + 2z + 5) \)

\( = (z + 1)(z^2 + 2z + 1 + 4) \)

\( = (z + 1)(z + 1)^2 - (2i)^2 \)

\( = (z + 1)(z + 1 - 2i)(z + 1 + 2i) \)

If \( P(z) = 0 \), then \( z = -1 \), \( z = -1 + 2i \), or \( z = -1 - 2i \).

Example 5

Solve the equation over \(\mathbb{C} \) :\( z^3 - iz^2 - 4z + 4i = 0 \).

Solution:

Factorise by grouping:

\( z^3 - iz^2 - 4z + 4i \)

\( = z^2(z - i) - 4(z - i) \)

\( = (z - i)(z^2 - 4) \)

\( = (z - i)(z - 2)(z + 2) \)

\( z^3 - iz^2 - 4z + 4i = 0 \)

\( (z - i)(z - 2)(z + 2) = 0 \)

Therefore, the solutions are:

\( z = i \), \( z = 2 \) or \( z = -2 \)

Example 6

Find the value of \(P\) given that: \(z + 2\) is a factor of \(z^{3} + 3z^2 + pz + 12\).

Solution

To find the value of \( p \) given that \( z + 2 \) is a factor of \( z^3 + 3z^2 + pz + 12 \), we can perform polynomial division. If \( z + 2 \) is a factor, the remainder of the division should be zero.

Let's perform the polynomial division: \[ \begin{array}{c|c} z + 2 & z^3 + 3z^2 + pz + 12 \\ \hline & - (z^3 + 2z^2) \\ & \underline{\phantom{-(z^2 + 2z)}} \\ & z^2 + pz + 12 \\ & - (z^2 + 2z) \\ & \underline{\phantom{-(pz + 2p)}} \\ & pz + 12 + 2z \\ & - (pz + 2p) \\ & \underline{\phantom{-12 + 2p}} \\ & 12 - 2p \\ \end{array} \]

Since the remainder is \( 12 - 2p \), and we want the remainder to be zero, we set \( 12 - 2p = 0 \) and solve for \( p \): \[ \begin{aligned} 12 - 2p &= 0 \\ 2p &= 12 \\ p &= \frac{12}{2} \\ p &= 6 \end{aligned} \]

Therefore, the value of \( p \) is 6.

Exercise &&1&& (&&1&& Question)

Factorise \(z^{3} - iz^{2} - 4z + 4i\).

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Exercise &&2&& (&&1&& Question)

Factorize: \(P(z) = z^{4}-16\).

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Exercise &&3&& (&&1&& Question)

What are the roots of a complex polynomial?

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Exercise &&4&& (&&1&& Question)

Solve the equation over \(\mathbb{C} \) :  \( z^2 + 64 = 0 \).

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Exercise &&5&& (&&1&& Question)

Find the value of \(P\) given that: \(z - i\) is a factor of \(z^{3} + pz^2 + z - 4\).

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