Problem 10 - Advanced Engineering Mathematics

Question

Express the given quantity in the form \(a+i b\). \(\sinh \left(\frac{3 \pi}{2} i\right)\)

Step-by-Step Solution

Answer

The expression is \(0 - i\), or \(-i\).

Step 1: Express Hyperbolic Sine in Terms of Exponentials

The hyperbolic sine function can be expressed in terms of exponential functions: \[\sinh(z) = \frac{e^z - e^{-z}}{2}\]Given the problem, set \(z = \frac{3\pi}{2}i\).

Step 2: Substitute and Simplify the Exponentials

Substitute \(z = \frac{3\pi}{2}i\) in the expression:\[\sinh\left(\frac{3\pi}{2}i\right) = \frac{e^{\frac{3\pi}{2}i} - e^{-\frac{3\pi}{2}i}}{2}\]Using Euler's formula \(e^{i\theta} = \cos(\theta) + i\sin(\theta)\), evaluate the exponentials:- \(e^{\frac{3\pi}{2}i} = \cos\left(\frac{3\pi}{2}\right) + i\sin\left(\frac{3\pi}{2}\right) = 0 - i = -i\)- \(e^{-\frac{3\pi}{2}i} = \cos\left(-\frac{3\pi}{2}\right) + i\sin\left(-\frac{3\pi}{2}\right) = 0 + i = i\)

Step 3: Perform the Final Calculation

Plug the evaluated exponentials back into the expression:\[\sinh\left(\frac{3\pi}{2}i\right) = \frac{-i - i}{2} = \frac{-2i}{2} = -i\]

Step 4: Express in the Form \(a + ib\)

The expression \(-i\) can be written as \(0 + (-1)i\), which is of the form \(a + ib\) with \(a = 0\) and \(b = -1\).

Key Concepts

Hyperbolic Functions

Hyperbolic functions are analogs of trigonometric functions, but they are defined using exponential functions. A key hyperbolic function is the hyperbolic sine, denoted as \( \sinh(x) \). It is defined by the formula:
\[ \sinh(x) = \frac{e^x - e^{-x}}{2} \]
This formula showcases how the hyperbolic sine function is related to exponential functions. Unlike trigonometric functions, hyperbolic functions are often used in calculus and complex analysis settings.

• Trigonometric functions circle around angles on the unit circle.
• Hyperbolic functions describe hyperbolas, which are curves on a different plane.

In contexts involving complex numbers, these functions reveal intriguing properties and help solve problems involving exponential growth or decay.

Euler's Formula

Euler's formula is a cornerstone concept in complex analysis and relates complex exponentials to trigonometry. It can be expressed as:
\( e^{i\theta} = \cos(\theta) + i\sin(\theta) \)
This formula indicates how complex exponentials map into the unit circle in the complex plane by combining both real and imaginary components. For example:

• The angle \( \theta \) determines the position on the circle.
• \( \cos(\theta) \) and \( i\sin(\theta) \) effectively describe the coordinates on this circle.

Euler's formula is invaluable when working with complex exponents and evaluating complex-valued functions, such as in the original exercise where it helps simplify hyperbolic functions of imaginary arguments.

Exponential Functions

Exponential functions are critical not just in pure mathematics but in various scientific fields. They typically have the form \( e^x \), where \( e \) is the base of the natural logarithm and approximately equals 2.71828.

• They model continuous growth or decay processes.
• They simplify expressions containing rates of change.

In complex analysis, exponential functions extend to include complex numbers, like \( e^{ix} \), which connect directly to Euler's formula. These functions, therefore, bridge the gap between algebraic expressions and geometric interpretations on the complex plane. Simplifying complex exponentials often involves expressing them in terms of coefficients of \( a + ib \), revealing both real and imaginary parts.

Complex Numbers

Complex numbers extend real numbers to solve equations that do not have real solutions, such as \( x^2 + 1 = 0 \). Denoted as \(a + ib\), where \(a\) and \(b\) are real numbers and \(i\) is the imaginary unit \(i^2 = -1\).

• The real part is \(a\).
• The imaginary part is \(b\).

These components allow complex numbers to be represented on the complex plane, where the horizontal axis is the real part and the vertical axis is the imaginary part. In the exercise, simplifying to \(-i\) as \(0 + (-1)i\) directly uses the real and imaginary definitions, illustrating how these concepts underpin the solution of complex-valued expressions.