Problem 90 - Algebra A Combined Function

Question

Write each expression in the form \(a+b i .\) $$ i^{8}-i^{7} $$

Step-by-Step Solution

Answer

The expression is equal to \( 1 + i \).

Step 1: Simplify Powers of i

Recall that the imaginary unit i has a cyclical pattern every four powers: \[i = i, \quad i^2 = -1, \quad i^3 = -i, \quad i^4 = 1\]Repeat this cycle for higher powers. For this problem, determine what \(i^8\) and \(i^7\) are based on this cycle.

Step 2: Evaluate i^8

Since \(i^4 = 1\), repeating this gives \(i^8 = (i^4)^2 = 1^2 = 1\). Now we have the simplified form for \(i^8\).

Step 3: Evaluate i^7

Using the cycle, first find \(i^7\) as \(i^7 = i^4 \cdot i^3\). Since \(i^4 = 1\) and \(i^3 = -i\), it follows that \(i^7 = 1 \cdot (-i) = -i\).

Step 4: Subtract Powers of i

Substitute the evaluated powers into the original expression: \[i^8 - i^7 = 1 - (-i)\] which simplifies to\[1 + i\].

Step 5: Write in a + bi Form

The result \(1 + i\) is already in the form \(a + bi\), where \(a = 1\) and \(b = 1\).

Key Concepts

Imaginary Unit

The imaginary unit, often denoted as \(i\), is the foundation of imaginary and complex numbers. In mathematics, \(i\) is defined as the square root of \(-1\). This definition might seem odd because there is no real number whose square is negative. But in the realm of complex numbers, \(i^2 = -1\).

• Role in Complex Numbers: \(i\) is the imaginary part of complex numbers, which are expressed in the form \(a + bi\), where \(a\) and \(b\) are real numbers.
• Utility: Although imaginary numbers may not exist on the regular number line, they are crucial for solving equations like \(x^2 + 1 = 0\), which do not have real solutions.

The concept of the imaginary unit allows us to extend the number system, enabling the solution of a broader set of mathematical problems.

Powers of i

Calculating the powers of the imaginary unit \(i\) is simpler than it may initially appear due to a repeating pattern. Here's how it works:

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

Notice the cycle repeats every four powers. Therefore, any power of \(i\) can be reduced by identifying the remainder when the power is divided by 4. This means for any integer power \(n\):

• If \(n \mod 4 = 0\), then \(i^n = 1\)
• If \(n \mod 4 = 1\), then \(i^n = i\)
• If \(n \mod 4 = 2\), then \(i^n = -1\)
• If \(n \mod 4 = 3\), then \(i^n = -i\)

Understanding this cyclical nature allows for easy computation of complex numbers involving powers of \(i\).

Cyclical Pattern of i

Once you grasp the cyclical pattern of the powers of \(i\), working with complex problems becomes significantly easier. The pattern is: \(i, -1, -i, 1\). This sequence repeats indefinitely.
This repetitive cycle helps simplify expressions with high powers of \(i\) without laborious computation:

• For example, \(i^7\) can be broken down into \(i^4\) and \(i^3\), resulting in \(1 \cdot (-i) = -i\).
• Similarly, \(i^8\) simplifies to \((i^4)^2 = 1^2 = 1\).

Learning to recognize and apply this cycle means you can manage and solve complex problems faster, giving you more confidence in handling complex numbers.

Complex Arithmetic

Complex arithmetic follows specific rules that involve both the real and imaginary parts of complex numbers. Complex numbers are written as \(a + bi\), where \(a\) is the real part and \(b\) is the coefficient of the imaginary part.

• Addition: Combine like terms: add real parts, add imaginary parts \((a + bi) + (c + di) = (a+c) + (b+d)i\).
• Subtraction: Similar to addition, subtract real and imaginary parts separately \((a + bi) - (c + di) = (a-c) + (b-d)i\).
• Multiplication: Use the distributive property \((a + bi) \cdot (c + di) = (ac-bd) + (ad+bc)i\).

Remember, multiplying two imaginary numbers results in a real number, because \(i^2 = -1\). Simplifying expressions is key, as demonstrated in our example, which ended up with \(1 + i\) after complex arithmetic operations on powers of \(i\). These foundational arithmetic rules allow us to handle complex numbers much like real numbers.