Problem 28
Question
Use (4)-(8) to find the derivative \(f^{\prime}(z)\) for the given function. $$ f(z)=5 z^{4}-i z^{3}+(8-i) z^{2}-6 i $$
Step-by-Step Solution
Verified Answer
The derivative \( f^{\prime}(z) = 20z^3 - 3iz^2 + (16-2i)z \).
1Step 1: Identify the Function
The given function is a polynomial function expressed as: \[ f(z) = 5z^4 - iz^3 + (8-i)z^2 - 6i \] We need to find the derivative of this function.
2Step 2: Apply the Power Rule
The Power Rule states that the derivative of \( z^n \) is \( nz^{n-1} \). Using this, we find the derivative of each term in the polynomial individually.
3Step 3: Differentiate Each Term
1. Differentiate the first term: \( \frac{d}{dz}(5z^4) = 20z^3 \). 2. Differentiate the second term: \( \frac{d}{dz}(-iz^3) = -3iz^2 \). 3. Differentiate the third term: \( \frac{d}{dz}((8-i)z^2) = 2(8-i)z = (16 - 2i)z \). 4. Differentiate the constant term: \( \frac{d}{dz}(-6i) = 0 \) because the derivative of a constant is zero.
4Step 4: Combine the Derivatives
Now combine all the differentiated terms to find the derivative of the original function. \[ f^{\prime}(z) = 20z^3 - 3iz^2 + (16-2i)z \].
Key Concepts
Polynomial FunctionsPower Rule in CalculusComplex Numbers in Calculus
Polynomial Functions
Polynomial functions are mathematical expressions involving a sum of powers of variable(s), each multiplied by a coefficient. In simpler terms, a polynomial is like a puzzle made up from multiple "pieces" called terms. Each term might look something like "5z^4" or "-iz^3". These pieces are glued together by addition or subtraction.
In the context of the problem, our polynomial function is:
Polynomial functions are simple yet powerful, forming the backbone of many mathematical concepts and real-life applications such as physics and engineering.
Understanding these functions aids in grasping how they can change and interact, which is the first step to differentiating them.
In the context of the problem, our polynomial function is:
f(z) = 5z^4 - iz^3 + (8-i)z^2 - 6i
z is called a "degree". For example, the degree of 5z^4 is 4. The degree of a polynomial is determined by the highest power of the variable present. Polynomial functions are simple yet powerful, forming the backbone of many mathematical concepts and real-life applications such as physics and engineering.
Understanding these functions aids in grasping how they can change and interact, which is the first step to differentiating them.
Power Rule in Calculus
The Power Rule is a fundamental tool in calculus, particularly in differential calculus, making the process of finding derivatives much more accessible. This rule helps us figure out how functions change, which is crucial for tasks like predicting trends and optimizing outcomes.
Simply put, if you have a term in a function such as
In the derivative calculation of the polynomial given, we applied the Power Rule to each term separately:
Simply put, if you have a term in a function such as
z^n, the Power Rule says the derivative is nz^{n-1}. You bring down the power in front of the term and then subtract one from the power. It's like a quick trick to find out the rate at which your function changes. In the derivative calculation of the polynomial given, we applied the Power Rule to each term separately:
- For
5z^4, we get20z^3. - For
-iz^3, we get-3iz^2. - For
(8-i)z^2, the derivative is(16-2i)z. - The constant term
-6igives us 0, as constants vanish in derivatives.
Complex Numbers in Calculus
Complex numbers introduce a fascinating dimension to calculus, enabling us to solve problems that are unsolvable with just real numbers. A complex number is expressed as
In our exercise, we deal with coefficients that include
a + bi, where a and b are real numbers, and i is the imaginary unit, satisfying i^2 = -1. In our exercise, we deal with coefficients that include
i. The derivative still follows the same rules, but now you're working in a "bigger playground" called the complex plane. Complex numbers need special attention because they offer more comprehensive solutions.
- The polynomials include terms like
-iz^3and(8-i)z^2, showcasing the elegance of combining real and imaginary parts. - Differentiating them involves multiplying the imaginary unit just like any constant.
Other exercises in this chapter
Problem 28
Express \(\ln z\) in the form \(a+i b\). \(z=-\sqrt{3}+i\)
View solution Problem 28
Verify that the given function \(u\) is harmonic. Find \(v\), the harmonic conjugate function of \(u\). Form the corresponding analytic function \(f(z)=u+i v .\
View solution Problem 28
$$ \text { Prove that } \cos ^{2} z+\sin ^{2} z=1 $$
View solution Problem 28
In Problems 23-28, express \(\ln z\) in the form \(a+i b\). $$ z=-\sqrt{3}+i $$
View solution