Problem 13 - Advanced Engineering Mathematics

Question

$$\int_{\pi}^{\pi+2 i} \sin \frac{z}{2} d z=-\left.2 \cos \frac{z}{2}\right|_{\pi} ^{\pi+2 i}=-2\left[\cos \left(\frac{\pi}{2}+i\right)-\cos \frac{\pi}{2}\right]=2 i \sin \frac{\pi}{2} \sinh 1=2.3504 i$$

Step-by-Step Solution

Answer

The value of the integral is \( 2.3504i \).

Step 1: Understanding the Integral

The given integral is \( \int_{\pi}^{\pi+2i} \sin \frac{z}{2} \, dz \). This means we are integrating the function \( \sin \frac{z}{2} \) with respect to \( z \) from \( z = \pi \) to \( z = \pi + 2i \).

Step 2: Finding the Antiderivative

The antiderivative of \( \sin \frac{z}{2} \) is \( -2 \cos \frac{z}{2} \). This is because the derivative of \( -2 \cos \frac{z}{2} \) with respect to \( z \) gives \( \sin \frac{z}{2} \).

Step 3: Evaluating the Antiderivative

Apply the limits of the definite integral to the antiderivative: \( -2 \bigg[ \cos \frac{z}{2} \bigg]_{\pi}^{\pi + 2i} \). This evaluates to \( -2 \left[ \cos \left( \frac{\pi}{2} + i \right) - \cos \frac{\pi}{2} \right] \).

Step 4: Simplifying the Cosine Terms

\( \cos \frac{\pi}{2} \) simplifies to 0 since the cosine of \( \frac{\pi}{2} \) is 0. So, the expression becomes:\( -2 \times \cos \left( \frac{\pi}{2} + i \right) \).

Step 5: Evaluating \( \cos \left( \frac{\pi}{2} + i \right) \)

Using the identity for \( \cos(a + ib) = \cos a \cosh b - i \sin a \sinh b \), substitute \( a = \frac{\pi}{2} \) and \( b = 1 \): \( \cos \frac{\pi}{2} \cosh 1 - i \sin \frac{\pi}{2} \sinh 1 \). Since \( \cos \frac{\pi}{2} = 0 \) and \( \sin \frac{\pi}{2} = 1 \), it simplifies to \( -i \sinh 1 \).

Step 6: Completing the Evaluation

Now the integral evaluates to:\( -2 (-i \sinh 1) = 2i \sinh 1 \). The numerical value of \( \sinh 1 \) is approximately 1.1752, thus the integral simplifies to \( 2 \times 1.1752i = 2.3504i \).

Step 7: Final Answer

The value of the integral is \( 2.3504i \).

Key Concepts

Complex Numbers

Complex numbers are an extension of the real number system and are essential in advanced mathematics and physics. They are written in the form \(a + bi\), where \(a\) and \(b\) are real numbers, and \(i\) is the imaginary unit, defined by the property \(i^2 = -1\).

• **Real Part**: The real number \(a\) is called the real part of the complex number.
• **Imaginary Part**: The coefficient \(b\) is the imaginary part. When \(b = 0\), the complex number is wholly real.

Complex numbers enable the performance of arithmetic operations that involve the square roots of negative numbers. In applications like complex integration, which is part of complex analysis, these numbers are used to solve problems that extend beyond the limitations of real number analysis. We can plot complex numbers on a plane known as the complex plane, where the x-axis represents the real part and the y-axis represents the imaginary part. This visualization helps in understanding transformations within complex analysis, such as the movement from \(\pi\) to \(\pi + 2i\) in our integral example.

Definite Integral

In calculus, a definite integral represents the net area under a curve within a specified interval. It is a cornerstone concept that helps calculate areas, volumes, central points, and in many physics applications, it describes the accumulation of quantities. For a function \(f(z)\) defined over an interval from \(a\) to \(b\), the definite integral is noted as \(\int_{a}^{b} f(z) \, dz\). It is computed as the limit of a sum of the form \(\sum f(x_i) \, \Delta x_i\) as the partition of the interval becomes infinitely fine. The solution to our exercise involves the definite integral \(\int_{\pi}^{\pi + 2i}\), meaning that integration needs to consider the complex nature of limits. To evaluate this, we first find the antiderivative and then apply the limits. The process of evaluation requires transforming trigonometric functions, which in this case involve hyperbolic sine due to the imaginary upper limit, illustrating how definite integrals accommodate complex paths using the principles of analysis.

Hyperbolic Functions

Hyperbolic functions are analogs of trigonometric functions but correspond to a hyperbola instead of a circle. They appear frequently in complex calculus due to their properties that relate to exponential functions and their ability to transform into imaginary exponentials. Some key hyperbolic functions include:

• \(\sinh x = \frac{e^x - e^{-x}}{2}\)
• \(\cosh x = \frac{e^x + e^{-x}}{2}\)

In our integral, we encounter the hyperbolic sine \(\sinh(1)\). This function, combined with trigonometric counterparts, completes the solution by transforming it into an imaginary value that fits the complex plane dynamics. The expression \(2 \times 1.1752i = 2.3504i\) at the exercise conclusion results from combining the sine and hyperbolic sine, highlighting their role in evaluating complex integrals.