Problem 6 - Advanced Engineering Mathematics

Question

\(\int_{-2 i}^{1}\left(3 z^{2}-4 z+5 i\right) d z\)

Step-by-Step Solution

Answer

The integral is \(-19 + 13i\).

Step 1: Break Down the Integral

We need to integrate the function \(3z^2 - 4z + 5i\) with respect to \(z\) from \(-2i\) to \(1\). Identify this as a definite integral involving polynomial and constant terms.

Step 2: Integrate the Function

Integrate each term of the integrand independently. The integral of \(3z^2\) is \(z^3\), the integral of \(-4z\) is \(-2z^2\), and the integral of \(5i\) is \(5iz\). The antiderivative is thus \(z^3 - 2z^2 + 5iz + C\).

Step 3: Evaluate the Definite Integral

Evaluate the antiderivative from \(-2i\) to \(1\). Calculate \([1^3 - 2(1)^2 + 5i(1)] - [(-2i)^3 - 2(-2i)^2 + 5i(-2i)]\).

Step 4: Calculate Upper Limit Evaluation

Substitute \(z = 1\) in \(z^3 - 2z^2 + 5iz\). This results in \(1^3 - 2(1)^2 + 5i(1) = 1 - 2 + 5i = -1 + 5i\).

Step 5: Calculate Lower Limit Evaluation

Substitute \(z = -2i\) in \(z^3 - 2z^2 + 5iz\). Use \((-2i)^3 = -8i\) and \((-2i)^2 = -4\), giving \(-8i - 2(-4) + 5i(-2i) = -8i + 8 + 10 = 18 - 8i\).

Step 6: Subtract Antiderivative Values

Subtract the value at the lower limit from the value at the upper limit: \((-1 + 5i) - (18 - 8i) = -1 + 5i - 18 + 8i = -19 + 13i\).

Key Concepts

Definite Integral

A definite integral computes the accumulation of quantities, like area under a curve, between two specific limits. In the context of complex numbers, definite integration operates similarly to real numbers but now involving complex limits and integrands. Here, we evaluate the integral

• from a lower bound: \(-2i\),
• to an upper bound: \(1\).

The process involves finding an antiderivative of the integrand \(3z^2 - 4z + 5i\), and then calculating the difference

• between the antiderivative's value at upper limit,
• and its value at lower limit.

This subtraction yields the result of the definite integral, demonstrating how much the function accumulates over the given interval. This accumulation isn't just about area, especially in the complex plane, but a more nuanced concept of change across that interval.

Complex Numbers

Complex numbers extend real numbers, incorporating an imaginary unit \(i\), where \(i^2 = -1\).
In integration, they provide a way to handle integrations over paths in the complex plane. With complex limits, such as \(-2i\), you have intricate paths and directions the integration must consider.

• Complex numbers add depth to integration, opening paths not just horizontally.

Working with integrals in the complex plane requires treating both the real and imaginary parts of numbers separately during computation. For example:

• The real part, like \(-4z\), undergoes normal integration techniques.
• The imaginary parts, like \(5i\), hold steady for integration while considering imaginary coefficient effects.

Understanding each component's behavior, real or imaginary, is crucial to mastering complex integrals.

Polynomial Integration

Integrating polynomials is often the first step in evaluating more complex integrals. Each term in a polynomial function integrates based on its degree, providing a straightforward method to find antiderivatives. Consider the polynomial \(3z^2 - 4z + 5i\):

• The term \(3z^2\) integrates to \(z^3\), as you add 1 to the power and divide by the new power.
• The term \(-4z\) becomes \(-2z^2\), following the same principle.
• The constant \(5i\) transforms to \(5iz\), translating constant multiplication rules.

This antiderivative \(z^3 - 2z^2 + 5iz\) forms the basis to compute the definite integral over any range or path specified, such as between a complex lower and upper limit. Knowing how to integrate each part instinctively simplifies tackling more advanced problems. The building block is always understanding these basic transformations correctly.