Problem 13 - Advanced Engineering Mathematics

Question

\(\int_{\pi}^{\pi+2 i} \sin \frac{z}{2} d z\)

Step-by-Step Solution

Answer

The integral evaluates to approximately \\( 2.3504i \\\).

Step 1: Define the Integral Path

The integral given is a complex integral along a path from \( \pi \) to \( \pi + 2i \). This path can be considered as a straight line in the complex plane. The path can be parameterized as \( z(t) = \pi + 2it, \; t \; ext{in} \; [0,1] \).

Step 2: Differentiate Path with Respect to Parameter

Compute the derivative \( rac{dz}{dt} \) using the parameterization \( z(t) = \pi + 2it \). Since \( z(t) \) is linear, \( rac{dz}{dt} = 2i \).

Step 3: Substitute and Simplify the Integral

The integral becomes \[ \int_{0}^{1} \sin \frac{\pi + 2it}{2} (2i) \ dt. \] You should substitute \( rac{z}{2} \) as \( rac{\pi}{2} + it \), making it \( \sin \left( rac{\pi}{2} + it \right) \).

Step 4: Use the Exponential Form of Sine Function

Utilize the expression \( \sin(a+bi) = \frac{e^{i(a+bi)} - e^{-i(a+bi)}}{2i} \) to express \( \sin \left( rac{\pi}{2} + it \right) \) as \[ \sin \left( rac{\pi}{2} \right) \cosh(t) + i \cos \left( rac{\pi}{2} \right) \sinh(t). \]

Step 5: Evaluate the Integral

Since \( \sin \left( rac{\pi}{2} \right) = 1 \) and \( \cos \left( rac{\pi}{2} \right) = 0 \, \) the integral turns into \[ \int_{0}^{1} (2i) \cdot \cosh(t) \ dt = 2i \cdot \left[ \sinh(t) \right]_{0}^{1} = 2i \cdot \sinh(1). \] The result of this is \( 2i \sinh(1) \).

Step 6: Calculate \\( \\\\sinh(1) \\\\)

The hyperbolic sine function at 1 is calculated as \(\ ext{sinh}(1) = rac{e - e^{-1}}{2} \ ext{which is approximately 1.1752}. \) Thus, the integral evaluates to approximately \( 2i \cdot 1.1752 = 2.3504i \).

Key Concepts

Complex Integration

In complex analysis, one of the fundamental tasks is integrating functions along paths in the complex plane. These integrals, known as complex integrals, extend the familiar concept of integration from real-valued functions to complex-valued functions.
This involves integrating with respect to a complex variable, often along a specific contour or path. The exercise given here requires evaluating a complex integral along a specified path, which is a straight line from \( \pi \) to \( \pi + 2i \). By parameterizing the path appropriately, complex integration can be performed using techniques similar to those employed in the integration of real functions.

• **Path Parameterization:** Important for converting the integral into something more familiar. It gives us a simplified way to express the integration process over a specific path.
• **Derivative with Respect to Parameter:** Indicates how the complex path changes, a crucial step as it transforms the path into a function of a single real variable.

Once the integral is expressed as a function of a real parameter, it allows the application of familiar calculus principles to solve it.

Complex Plane

The complex plane is a unique mathematical landscape where complex numbers are graphed. Each point in this plane represents a complex number \( z = x + yi \), where \( x \) is the real part, and \( y \) is the imaginary part. Imagine it like a coordinate system where the x-axis denotes the real part and the y-axis denotes the imaginary coefficient.

• **Visualizing Complex Numbers:** Allows for easy representation of addition, subtraction, and combination of numbers via geometric concepts like vectors.
• **Paths in Complex Plane:** Complex integrals often involve integration along a path or contour, like the straight line from \( \pi \) to \( \pi + 2i \) in our example.

The imaginary unit \( i \) essentially rotates a real number by 90 degrees, adding depth to our understanding of the interactions between real and imaginary components. This powerful visualization tool helps in intuitively grasping concepts and solving complex integration problems.

Hyperbolic Functions

Hyperbolic functions, much like their circular counterparts (sine and cosine), are associated with a hyperbola rather than a circle. In complex analysis, these functions often come into play when dealing with exponential forms. The hyperbolic sine and cosine functions are defined as follows:

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

In the context of our integral, hyperbolic functions help simplify expressions involving complex numbers in exponential forms. Hyperbolic functions provide a clear method of describing growth in solutions, especially when conventional trigonometric functions would not suffice.
These functions prove especially useful when working in the complex plane, as they lend themselves well to simplification and transformation of complex integrals.

Exponential Form of Trigonometric Functions

The exponential form of trigonometric functions provides a powerful way to express complex sine and cosine. This form is very useful in simplifying complex expressions and solving integrals involving trigonometric functions.
For a complex number \( z = a + bi \), sine and cosine can be expressed using exponential functions:

• **Sine:** \( \sin(a + bi) = \frac{e^{i(a+bi)} - e^{-i(a+bi)}}{2i} \)
• **Cosine:** \( \cos(a + bi) = \frac{e^{i(a+bi)} + e^{-i(a+bi)}}{2} \)

In solving our original problem, this exponential form proved essential for rewriting the integral's expression. By doing this, we broke down complex trigonometric identities into manageable components. These transformations simplify the integration of complex paths and functions by reducing intricate expressions to familiar exponential terms, allowing straightforward integration steps.