Problem 16 - Advanced Engineering Mathematics

Question

$$\begin{aligned} \int_{i}^{1+\frac{\pi}{2} i} \sinh 3 z d z &=\left.\frac{1}{3} \cosh 3 z\right|_{i} ^{1+\frac{\pi}{2} i}=\frac{1}{3}\left[\cosh \left(3+\frac{3 \pi}{2} i\right)-\cosh 3 i\right] \\ &=\frac{1}{3}\left[\cosh 3 \cos \frac{3 \pi}{2}+i \sinh 3 \sin \frac{3 \pi}{2}-\cos 3\right]=-\frac{1}{3} \cos 3-\frac{1}{3} i \sinh 3=0.3300-3.3393 i \end{aligned}$$

Step-by-Step Solution

Answer

The integral evaluates to the complex number \( 0.3300 - 3.3393i \).

Step 1: Identify the Integral

The given problem is to evaluate the integral \( \int_{i}^{1+\frac{\pi}{2} i} \sinh 3z \, dz \). This is a definite integral of the hyperbolic sine function \( \sinh 3z \) over a path in the complex plane from \( z=i \) to \( z=1+\frac{\pi}{2}i \).

Step 2: Find the Antiderivative of the Integrand

The antiderivative of \( \sinh 3z \) can be found. Since the derivative of \( \cosh 3z \) is \( 3\sinh 3z \), the antiderivative is \( \frac{1}{3} \cosh 3z \).

Step 3: Evaluate the Antiderivative at the Bounds

Evaluate the antiderivative \( \frac{1}{3} \cosh 3z \) at the upper bound and lower bound of the integral: - At \( z = 1 + \frac{\pi}{2} i \), we have \( \frac{1}{3} \cosh(3 + \frac{3\pi}{2}i) \).- At \( z = i \), we have \( \frac{1}{3} \cosh 3i \).

Step 4: Substitute and Simplify

Substitute these into the integral formula \( \frac{1}{3}[\cosh(3 + \frac{3\pi}{2}i) - \cosh 3i] \).

Step 5: Evaluate Hyperbolic and Trigonometric Functions

Using the identities \( \cosh(a+bi) = \cosh a \cos b + i \sinh a \sin b \) and \( \cosh bi = \cos b \), we simplify:- \( \cosh(3 + \frac{3\pi}{2}i) = \cosh 3 \cos \frac{3\pi}{2} + i \sinh 3 \sin \frac{3\pi}{2} \).- \( \cosh 3i = \cos 3 \).

Step 6: Compute Final Result

Evaluate each trigonometric and hyperbolic function:- \( \cos \frac{3\pi}{2} = 0 \), \( \sin \frac{3\pi}{2} = -1 \)- Replace to find \( \frac{1}{3}(0 \cdot \cosh 3 + i \cdot \sinh 3 \cdot (-1) - \cos 3) \).- Simplify to get \( -\frac{1}{3} \cos 3 - \frac{1}{3} i \sinh 3 \).

Step 7: Express Final Answer

Compute the numerical values: \( \cosh 3 \approx 10.0677 \) and \( \sinh 3 \approx 10.0178 \) to express in the closest decimal form:- Use these to calculate and obtain the resultant complex number \( 0.3300 - 3.3393i \).

Key Concepts

Hyperbolic Functions

Hyperbolic functions like \( \sinh z \) and \( \cosh z \), are analogs of the trigonometric functions \( \sin z \) and \( \cos z \). They are important in both real and complex analysis. For complex numbers, these functions play a significant role when evaluating integrals over complex paths.

The hyperbolic sine function, \( \sinh z = \frac{e^z - e^{-z}}{2} \), relates closely to the exponential function, helping us simplify calculations. In this exercise, \( \sinh 3z \) is the key part of the integrand, and understanding its properties aids in solving the integral.

Recognizing hyperbolic function identities such as \( \cosh^2 z - \sinh^2 z = 1 \) can simplify complex integrals. These identities are somewhat parallel to trigonometric ones, which makes them versatile for different mathematical tasks.

Antiderivative

Finding the antiderivative is like reversing differentiation. It's the process needed to solve integrals. For this problem, identifying \( \frac{1}{3} \cosh 3z \) as the antiderivative of \( \sinh 3z \) is a crucial step. This choice relies on the derivative rule that \( \frac{d}{dz} \cosh z = \sinh z \).

In complex analysis, knowing antiderivatives of hyperbolic and trigonometric functions immensely aids in evaluating definite integrals. Here, it allows us to directly compute the integral across the specified bounds in the complex plane.

When you find antiderivatives, always double-check your result by differentiating. This ensures accuracy and acts as a good habit in solving calculus problems.

Complex Plane

The complex plane is like a graph for complex numbers. It has a real axis and an imaginary axis. Any complex number, like \( z = a + bi \), can be plotted on this plane, where \( a \) is the real part and \( b \) is the imaginary part.

In this context, definite integrals are evaluated over paths or curves in the complex plane. The integral from \( i \) to \( 1 + \frac{\pi}{2}i \) means we are integrating along a path traced from these points.

Complex integration is fundamental in fields such as engineering and physics, where complex numbers describe waves or oscillations. Understanding how to navigate the complex plane makes complex integrals manageable.

Trigonometric Identities

Trigonometric identities, like \( \cos \theta + i \sin \theta = e^{i\theta} \), are powerful tools. They relate angles and lengths in triangles to one another and help evaluate expressions like \( \cosh 3z \) in the complex domain.

In this exercise, trigonometric identities facilitated converting \( \cosh(3 + \frac{3\pi}{2}i) \) into simpler terms like \( \cosh 3 \cos \frac{3\pi}{2} + i \sinh 3 \sin \frac{3\pi}{2} \). This simplification is key to computing the result efficiently.

Memorizing and understanding core trigonometric identities and their extensions to the complex plane unlocks easier calculations and richer insights into the behavior of complex functions.