Problem 4 - Advanced Engineering Mathematics

Question

Write \(f(z)=z^{4}-16=\left(z^{2}-4\right)\left(z^{2}+4\right)=(z-2)(z+2)(z-2 i)(z+2 i)\) to see that \(2,-2,2 i,\) and \(-2 i\) are zeros of \(f .\) Now \(f^{\prime}(z)=4 z^{3}\) and \(f^{\prime}(2) \neq 0, f^{\prime}(-2) \neq 0, f^{\prime}(2 i) \neq 0,\) and \(f^{\prime}(-2 i) \neq 0 .\) This indicates that each zero is of order one.

Step-by-Step Solution

Answer

Each zero is of order one because the derivative is non-zero at each zero.

Step 1: Factor the Polynomial

The given polynomial is \( f(z) = z^4 - 16 \). This can be factored into two quadratic expressions as \( f(z) = (z^2 - 4)(z^2 + 4) \). Each quadratic can be further factored into linear factors: \( z^2 - 4 = (z - 2)(z + 2) \) and \( z^2 + 4 = (z - 2i)(z + 2i) \). Thus, \( f(z) = (z - 2)(z + 2)(z - 2i)(z + 2i) \).

Step 2: Identify the Zeros

The factored form \( (z - 2)(z + 2)(z - 2i)(z + 2i) \) indicates the zeros of the polynomial. These zeros are \( z = 2, -2, 2i, \text{ and } -2i \).

Step 3: Compute the Derivative

The derivative of the polynomial \( f(z) = z^4 - 16 \) is found using the power rule. Thus, \( f'(z) = 4z^3 \).

Step 4: Evaluate the Derivative at Zeros

Let's evaluate \( f'(z) = 4z^3 \) at each zero. For \( z = 2 \), \( f'(2) = 4(2)^3 = 32 \). For \( z = -2 \), \( f'(-2) = 4(-2)^3 = -32 \). For \( z = 2i \), \( f'(2i) = 4(2i)^3 = -32i \). For \( z = -2i \), \( f'(-2i) = 4(-2i)^3 = 32i \). In all cases, \( f'(z) eq 0 \).

Step 5: Determine the Order of Zeros

Since \( f'(z) eq 0 \) at each of the zeros \( 2, -2, 2i, \text{ and } -2i \), it indicates that each zero is a simple zero. Therefore, each zero is of order one.

Key Concepts

Zeros of a Polynomial

Understanding the zeros of a polynomial is crucial in complex analysis. Zeros are the values of the variable that make the polynomial equal to zero. For the polynomial \(f(z) = z^4 - 16\), finding its zeros involves setting the polynomial equal to zero and solving for \(z\). This means:

• Setting \(z^4 - 16 = 0\)
• Factoring it into its simpler forms until the polynomial is expressed in terms of its linear factors.

For \(f(z)\), the zeros can be easily spotted once you factor it as \((z - 2)(z + 2)(z - 2i)(z + 2i)\). The corresponding zeros are \(z = 2, -2, 2i,\) and \(-2i\). These are the points where the polynomial intersects the complex plane, effectively giving zero value.

Polynomial Factorization

Polynomial factorization is the process of expressing a polynomial as the product of its factors. More simply, it involves breaking down a polynomial into simpler polynomials that can be multiplied to get the original polynomial.
To factor \(f(z) = z^4 - 16\), we first observe it as a difference of squares: \(z^4 - 16\) can be rewritten as \((z^2)^2 - 4^2\). Using the formula for differences of squares, \(a^2 - b^2 = (a - b)(a + b)\), we can factor it to:

• \((z^2 - 4)(z^2 + 4)\)

Further factorizing \(z^2 - 4\) and \(z^2 + 4\), we use the basic algebraic identity: \(x^2 - y^2 = (x - y)(x + y)\). Thus, we go one step further:

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

Ultimately, giving \((z - 2)(z + 2)(z - 2i)(z + 2i)\) as the fully factored form of the polynomial.

Complex Numbers

Complex numbers are essential in mathematics, especially when dealing with polynomial equations that don’t have real roots. A Complex number has a real part and an imaginary part, written in the form \(a + bi\), where \(i\) is the imaginary unit satisfying \(i^2 = -1\).
In the polynomial \(f(z) = z^4 - 16\), zeros such as \(2i\) and \(-2i\) are purely imaginary. This highlights the power of complex numbers in solving equations that are otherwise unsolvable using only real numbers. Through such numbers, we can better grasp the full scope of mathematical functions, especially in polynomial expressions.

Derivative Evaluation

Evaluating the derivative of a polynomial helps us understand the behavior and different characteristics of the polynomial’s graph. The derivative, \(f'(z)\), unveils the rate at which our polynomial changes as \(z\) changes.
For \(f(z) = z^4 - 16\), its derivative can be calculated using the power rule, which provides us with \(f'(z) = 4z^3\). Evaluating this derivative at the polynomial's zeros \(2, -2, 2i,\) and \(-2i\) reveals their respective derivatives:

• For \(z = 2\), \(f'(2) = 32\)
• For \(z = -2\), \(f'(-2) = -32\)
• For \(z = 2i\), \(f'(2i) = -32i\)
• For \(z = -2i\), \(f'(-2i) = 32i\)

Interestingly, in each case, \(f'(z)\) does not equal zero, ensuring that these zeros are simple and of order one. Such derivative evaluations are fundamental to determining the nature and characteristics of each zero.