Problem 11 - Complex Analysis for Mathematics and Engineering

Question

Evaluate \(\int_{C} z^{2} d z\), where \(C\) is the line segment from 1 to \(1+i\).

Step-by-Step Solution

Answer

The evaluated integral is \(-1 + \frac{4i}{3}\).

Step 1: Parameterize the curve

The line segment from 1 to \(1+i\) can be parameterized using the parameter \(t\), where \(0 \leq t \leq 1\). The parameterization is given by \(z(t) = 1 + ti\).

Step 2: Find the derivative

To find \(\frac{dz}{dt}\), differentiate the parameterization: \(z(t) = 1 + ti\), hence \(\frac{dz}{dt} = i\).

Step 3: Substitute into the integral

Substitute \(z(t) = 1 + ti\) and \(\frac{dz}{dt} = i\) into the integral. Our integral becomes: \[\int_{0}^{1} (1 + ti)^{2} i \, dt.\]

Step 4: Expand and simplify

Expand \((1 + ti)^{2}\) to get \(1 + 2ti + (ti)^{2} = 1 + 2ti - t^{2}\). So, the integral becomes: \[\int_{0}^{1} (1 + 2ti - t^{2}) i \, dt.\]

Step 5: Distribute and separate the integral

Distribute \(i\) inside the integral:\[\int_{0}^{1} (i + 2ti^2 - t^{2}i) \, dt = \int_{0}^{1} (i - 2t + t^{2}i) \, dt.\]This can be separated into three integrals:\[\int_{0}^{1} i \, dt - \int_{0}^{1} 2t \, dt + \int_{0}^{1} t^{2}i \, dt.\]

Step 6: Evaluate each integral

Evaluate each integral:1. \(\int_{0}^{1} i \, dt = [it]_{0}^{1} = i\).2. \(\int_{0}^{1} 2t \, dt = [t^{2}]_{0}^{1} = 1\).3. \(\int_{0}^{1} t^{2}i \, dt = \int_{0}^{1} it^{2} \, dt = i\left[\frac{t^{3}}{3}\right]_{0}^{1} = \frac{i}{3}\).

Step 7: Combine the results

Sum the results of the three integrals: \[i - 1 + \frac{i}{3} = i + \frac{i}{3} - 1 = \frac{4i}{3} - 1.\] This simplifies to the complex number: \(-1 + \frac{4i}{3}\).

Key Concepts

Contour Integration

Contour integration is a fabulous tool in complex analysis. It allows us to evaluate complex integrals over specific paths, known as contours, in the complex plane. By carefully choosing a path that fits well with the function you want to integrate, you can significantly simplify the evaluation process.

Here's why it's powerful:

• Contour integration uses the properties of analytic functions and the shape of the path to find the integral directly.
• Specially shaped contours can sometimes allow the use of tools like the Cauchy Integral Theorem or Residue Theorem, providing shortcuts to calculation.

When working with real functions, integration follows along intervals on the number line, but with complex functions, the path could be any curve in the complex plane, such as a circle or a line segment. In our example, the contour is simply a straight line from one complex number to another.

Parameterization of Curves

Parameterization is a way of describing a curve using a single parameter, often making it easier to handle in calculations. Here, the line segment from 1 to \(1+i\) is parameterized. By introducing a parameter \(t\), you can express points on the line as \(z(t) = 1 + ti\) with \(t\) ranging from 0 to 1.

Why parameterization?

• It simplifies the process by converting complex paths into a range over a single variable.
• Allows to easily compute derivatives which are crucial for integration.

In parameterizing, the start and end points of the contour are given specific values of the parameter, such as \(t=0\) for the start and \(t=1\) for the end. This method is fundamental in simplifying the integration over curves.

Complex Integrals

In complex analysis, integrals can be taken along paths in the complex plane, leading to what we call complex integrals. They look similar to real integrals, but with an additional layer of computation due to complex numbers.

Key aspects of complex integrals include:

• The path of integration significantly affects the result, unlike in real integrals where paths are simply intervals.
• They involve both real and imaginary components, which must be handled separately or through combined parameterization.

In our example, the integral \(\int_{C} z^{2} dz\) turned into a real integral after parameterization. This transformation was essential for applying traditional calculus techniques to evaluate it.

Differential Calculus

Differential calculus is the branch of mathematics dealing with how functions change. In the context of complex analysis, it often involves computing derivatives of functions concerning a complex variable.

Why use differentiations?

• To find the rate of change of a function, which is crucial for understanding its behavior along a path.
• In integrals, the differential \(dz\) is expressed in terms of a real parameter, requiring differentiation.

For our problem, differentiating \(z(t) = 1 + ti\) with respect to \(t\) gave \(\frac{dz}{dt} = i\). This derivative was used to transform the original complex integral into a standard form, making it solvable by conventional integration techniques we utilize in calculus.