Problem 4 - Advanced Engineering Mathematics

Question

In Problems 1-26, write the given number in the form \(a+i b\). $$ i^{11} $$

Step-by-Step Solution

Answer

The number in the form \(a + ib\) is \(-i\).

Step 1: Understanding Powers of i

The imaginary unit, denoted as \(i\), is defined as \(i = \sqrt{-1}\). Powers of \(i\) cycle in a sequence: \(i^1 = i\), \(i^2 = -1\), \(i^3 = -i\), and \(i^4 = 1\). This pattern repeats every four powers.

Step 2: Simplifying i^11

Since the powers of \(i\) repeat every four, we can simplify \(i^{11}\) by finding the remainder of 11 divided by 4. The remainder will tell us which power in the cycle \(i, -1, -i, 1\) to use.

Step 3: Finding the Remainder

We calculate \(11 \div 4\) to find the remainder: 11 divided by 4 is 2 with a remainder of 3. Therefore, \(i^{11} \equiv i^3\).

Step 4: Using the Cycle to Determine the Result

From the established cycle, \(i^3 = -i\), thus \(i^{11} = -i\).

Step 5: Writing in Standard Form

The standard form of a complex number is \(a + ib\). Since \(i^{11} = -i\), this can be expressed in the standard form as \(0 + (-1)i\), which simplifies to \(-i\).

Key Concepts

Imaginary Unit

The imaginary unit, represented by the symbol \(i\), is a fundamental concept in complex numbers. Its defining feature is that \(i\) is equal to the square root of \(-1\). This might seem confusing at first because we know that the square root of any negative number isn't a real number. That's why \(i\) is called 'imaginary'.

• \(i = \sqrt{-1}\)
• The introduction of \(i\) allows us to extend our number system beyond real numbers.

Understanding \(i\) is crucial for more advanced topics in mathematics, like working with complex numbers, which combine real and imaginary parts.

Powers of i

Powers of the imaginary unit \(i\) follow a predictable cycle. This cycle repeats every four powers, which makes calculating powers of \(i\) much simpler:

• \(i^1 = i\)
• \(i^2 = -1\)
• \(i^3 = -i\)
• \(i^4 = 1\)

After \(i^4\), the cycle repeats again, so \(i^5\) is the same as \(i^1\), \(i^6\) like \(i^2\), and so on. This cyclic nature is a useful property when simplifying expressions that involve higher powers of \(i\). For example, to simplify \(i^{11}\), observe the remainder when 11 is divided by 4 is 3. Therefore, \(i^{11} = i^3 = -i\).

Complex Number Form

A complex number is a number that has both a real and an imaginary part. It's written in the form \(a + ib\), where \(a\) and \(b\) are real numbers. Here, \(a\) is the real component, and \(b\) is the coefficient of the imaginary component \(i\).

• For \(a=0\) and \(b=1\), the complex number is purely imaginary (like \(i\)).
• For \(a=1\) and \(b=0\), the complex number is purely real (like any real number).

In our example, \(i^{11} = -i\), can be expressed as \(0 + (-1)i\) in complex number form, meaning it is purely imaginary with no real part.

Simplifying Expressions

Simplifying expressions involving complex numbers, particularly those with powers of \(i\), involves recognizing and utilizing its cyclical power property. Let's break down the steps:

• Identify the power of \(i\) being evaluated, like \(i^{11}\).
• Divide the power by 4 and determine the remainder.
• Use the remainder to simplify the expression, knowing the cycle is \(i, -1, -i, 1\).
• Express the final result in \(a + ib\) form.

Thus, by finding that \(i^{11} = -i\), we express it as \(-i\) or \(0 - i\). By repeating these steps, even more complex expressions involving \(i\) become easier to handle.