Problem 20 - Advanced Engineering Mathematics

Question

\(\int_{C}\left(4+3 z^{2}+2 z+1\right) d z ; C\) is the line segment from 0 to \(2 i\)

Step-by-Step Solution

Answer

The integral evaluates to \(2i - 4\).

Step 1: Parameterize the Curve

The line segment from 0 to \(2i\) can be parameterized using \(z(t) = 2it\), where \(t\) ranges from 0 to 1. This means that as \(t\) moves from 0 to 1, \(z\) moves from 0 to \(2i\).

Step 2: Differentiate the Parameterization

Differentiate \(z(t) = 2it\) with respect to \(t\) to get \(\frac{dz}{dt} = 2i\).

Step 3: Substitute and Set Up the Integral

Substitute \(z(t)\) and \(\frac{dz}{dt}\) into the integral: \[\int_{0}^{1} \left(4 + 3(2it)^2 + 2(2it) + 1\right) (2i) dt\].Simplify to get\[\int_{0}^{1} \,\left(4 + 3(-4t^2) + 4it + 1\right) (2i) \, dt\].

Step 4: Simplify the Expression

Simplify the expression inside the integral:\[4 + 3(-4t^2) + 4it + 1 = 5 - 12t^2 + 4it\].Thus, the integral becomes:\[\int_{0}^{1} (5 - 12t^2 + 4it)(2i) \, dt\].

Step 5: Expand and Integrate

Expand the product:\[\int_{0}^{1} (10i - 24it^2 - 8t) dt\].Integrate each term separately from 0 to 1:\[\int_{0}^{1} 10i \, dt - \int_{0}^{1} 24it^2 \, dt - \int_{0}^{1} 8t \, dt\].This results in:\[[10it]_{0}^{1} - [8it^3]_{0}^{1} - [4t^2]_{0}^{1} = 10i - 8i - 4.\]

Step 6: Final Step: Calculate the Result

Calculate the final result by evaluating:\[10i - 8i - 4 = 2i - 4.\]The final answer is \(2i - 4\).

Key Concepts

Line Parameterization

When you're working with complex functions and need to integrate along a specific path, line parameterization is your first step. Think of parameterization as setting a roadmap for your path.
For the exercise, we're looking at integrating over a line segment from 0 to \(2i\). In simpler terms, imagine a line stretching from point zero to a point at the top of the imaginary axis, at \(2i\).
To parameterize this line, we choose a variable, often \(t\), that helps us describe points on this line. In this exercise, we represent each point \(z\) on the path as \(z(t) = 2it\). As \(t\) moves from 0 to 1, \(z\) traces the entire line segment from 0 to \(2i\).

• At \(t = 0\), \(z = 0\); the starting point.
• At \(t = 1\), \(z = 2i\); the endpoint.

Line parameterization is crucial in defining how we set up the integral over the curve we are interested in. Once parameterized, differentiating with respect to \(t\) is a straightforward way to set up the integral for computation.

Curve Integration

Curve integration is a method used to integrate complex functions along a specified path or curve. In these scenarios, the usual path is not a straight line; it's more like tracing along a line or a curve in the complex plane.
For our problem, after parameterizing the line segment from 0 to \(2i\), we substitute back into the integral.
This substitution transforms the complex integral into one we can solve using standard methods. The integral is set up as follows: \[\int_{0}^{1} \left(4 + 3(2it)^2 + 2(2it) + 1\right) (2i) \, dt\]\The complex variable \(z\) and its derivative \(\frac{dz}{dt}\) are both expressed in terms of \(t\). Now, you deal with real integrals where the functions are comprised of terms involving the parameter \(t\). This makes calculation and simplification easier, allowing you to apply real analysis techniques right in the realm of complex numbers.

Definite Integral Evaluation

Evaluating the definite integral means calculating the exact area or "value" that the curve or region defined by the complex function covers in the given range.
Once you have set up the integral for the given parameterization, the next steps involve simplification and computation. For the given problem, after substituting and simplifying, you reach the integral:
\[\int_{0}^{1} (10i - 24it^2 - 8t) \, dt\]
This results in separate integrals:

• \(\int_{0}^{1} 10i \, dt\)
• \(-\int_{0}^{1} 24it^2 \, dt\)
• \(-\int_{0}^{1} 8t \, dt\)

You evaluate each integral independently. Evaluate considering the limits from 0 to 1. This structured approach results in simple arithmetic to achieve the final result, combining terms and arriving at the answer, \(2i - 4\).
Completing the integration process yields a tangible result from a visualized path, bridging together both theoretical and practical aspects of complex integration.