Problem 5 - Advanced Engineering Mathematics

Question

In Problems \(1-12\), express the given quantity in the form \(a+i b\). $$ \tan (i) $$

Step-by-Step Solution

Answer

\(\tan(i) = 0 + i \frac{e - e^{-1}}{e + e^{-1}}\)

Step 1: Understand the problem

We need to express \( \tan(i) \) in the form \( a + ib \), where \( a \) and \( b \) are real numbers.

Step 2: Recall the definition of tangent for complex numbers

The tangent of a complex number \( z \), represented as \( \tan(z) \), is given by the formula:\[\tan(z) = \frac{\sin(z)}{\cos(z)}\]

Step 3: Substitute the complex number into the formula

To find \( \tan(i) \), substitute \( z = i \) into the formula:\[\tan(i) = \frac{\sin(i)}{\cos(i)}\]

Step 4: Calculate \( \sin(i) \) and \( \cos(i) \)

Using the identities:- \( \sin(i) = \frac{e^i - e^{-i}}{2i} \) turns into \( \frac{e^{-1} - e}{2i} \)- \( \cos(i) = \frac{e^i + e^{-i}}{2} \) turns into \( \frac{e^{-1} + e}{2} \).Thus,\[\sin(i) = \frac{e^{-1} - e}{2i}, \quad \cos(i) = \frac{e^{-1} + e}{2}\]

Step 5: Divide \( \sin(i) \) by \( \cos(i) \)

To find \( \tan(i) \):\[\tan(i) = \frac{\frac{e^{-1} - e}{2i}}{\frac{e^{-1} + e}{2}} = \frac{e^{-1} - e}{i(e^{-1} + e)}\]

Step 6: Rationalize the expression

Multiply the numerator and the denominator by \( i \) to simplify:\[\tan(i) = \frac{(e^{-1} - e)i}{i^2(e^{-1} + e)} = \frac{(e^{-1} - e)i}{-(e^{-1} + e)}\]Thus,\[\tan(i) = \frac{(e^{-1} - e)i}{-(e^{-1} + e)} = \frac{e - e^{-1}}{e + e^{-1}}i\]So \( a = 0 \) and \( b = \frac{e - e^{-1}}{e + e^{-1}} \).

Step 7: Express in the form \( a + ib \)

Finally, express \( \tan(i) \) in the required form:\[\tan(i) = 0 + i \left( \frac{e - e^{-1}}{e + e^{-1}} \right)\]

Key Concepts

tangent function

The tangent function is a basic trigonometric function that can be extended to handle complex numbers. Tangent for real numbers is derived from a simple ratio: the sine divided by the cosine. Specifically, for any angle real angle \( \theta \), the tangent is expressed as:

• \( \tan(\theta) = \frac{\sin(\theta)}{\cos(\theta)} \)

Now, when we move into the realm of complex numbers, the definition remains similar. However, each trigonometric function must now be interpreted in terms of its complex exponential form. The expression for the tangent of a complex number \( z \) is:

• \( \tan(z) = \frac{\sin(z)}{\cos(z)} \)

This interpretation allows for the tangent function and all other trigonometric functions to be applied to anycomplex number, such as imaginary and real components. Hence, \( \tan(i) \) involves understanding sine and cosine for complex values.

imaginary unit

The imaginary unit is a fundamental concept in complex numbers, primarily represented by \( i \). Its defining property is that \( i^2 = -1 \). The use of this unit expands our number system beyond the real numbers, incorporatingelements like \( \sqrt{-1} \).Whenever a complex number appears, it is typically expressed in the form \( a + bi \), where \( a \) and \( b \) are real numbers, and \( i \) stands for the imaginary unit. The importance of \( i \) in mathematics opens up fields such as electrical engineering, quantum mechanics, and many others that deal with wave functions and oscillations.In the context of complex trigonometric functions, \( i \) participates actively in converting exponential functions into trigonometric identities via Euler's formula, which is important when simplifying and rationalizing expressions.

complex trigonometric functions

Complex trigonometric functions extend the familiar sine, cosine, and tangent functions to complex numbers. This involves redefining these functions using exponential functions, which helps in managing imaginary numbers and makes it easier to work with on complex planes.For example:

• The complex sine function is given by \( \sin(z) = \frac{e^{iz} - e^{-iz}}{2i} \).
• The complex cosine function is \( \cos(z) = \frac{e^{iz} + e^{-iz}}{2} \).

These expressions stem from Euler's formula, connecting trigonometric and exponential functions with complex arguments. They provide valuable identities for calculating functions like \( \tan(i) \), as shown in the problem where \( i \) is subbed into these formulas.This approach allows us to understand behaviors of trigonometric functions within the imaginary realm, makingit possible to express solutions in the form \( a + bi \), our crucial task.

rationalizing complex expressions

Rationalizing complex expressions involves transforming them into a form without complex numbers in the denominator. This is crucial for simplifying equations and achieving a standard form \( a + bi \), where it's easier to identify and compare its components.When working with expressions like \( \frac{e^{-1} - e}{i(e^{-1} + e)} \), we aim to eliminate the imaginary term in the denominator. This is done by multiplying the numerator and the denominator by the imaginary unit \( i \) to "move" the imaginary part out of the denominator.Let’s consider this step within the solved problem for \( \tan(i) \):

• Multiply the expression by \( i/i \) to get \( \frac{(e^{-1} - e)i}{i^2(e^{-1} + e)} \).
• Since \( i^2 = -1 \), we simplify to \( \frac{(e^{-1} - e)i}{-(e^{-1} + e)} \).

This new arrangement permits clearer analysis and interpretation, reducing complex expressions to real and imaginary components and making them manageable!