Problem 82
Question
Simplify. Write the result in the form \(a+b i .\) \(i^{100}\)
Step-by-Step Solution
Verified Answer
The simplified form is \(1 + 0i \).
1Step 1: Understand the Pattern of Powers of i
The imaginary unit, denoted as \( i \), has the property that \( i^2 = -1 \). The powers of \( i \) cycle every four terms: \( i^1 = i \), \( i^2 = -1 \), \( i^3 = -i \), and \( i^4 = 1 \). After \( i^4 \), the power pattern repeats.
2Step 2: Find the Equivalent Power of i in the Cycle
To simplify \( i^{100} \), we need to find which power it corresponds to in the cycle. We do this by dividing 100 by 4 to find the remainder: \( 100 \div 4 = 25 \) remainder \( 0 \). A remainder of 0 means the power corresponds to \( i^4 \), which equals to 1.
3Step 3: Express in the Form a + bi
Since \( i^{100} = (i^4)^{25} = 1^{25} = 1 \), we express this in the form of a complex number: \( 1 + 0i \). Here, \( a = 1 \) and \( b = 0 \).
Key Concepts
Powers of Imaginary UnitSimplification of Algebraic ExpressionsCyclic Patterns in Mathematics
Powers of Imaginary Unit
The imaginary unit, represented as \( i \), is crucial in complex number arithmetic. It is defined with the property that \( i^2 = -1 \). This unique property leads to a repeating cycle of powers, pivotal in simplifying expressions involving \( i \). Here's how it cycles:
This knowledge allows us to find the power of any large exponent involving \( i \) by determining its remainder when divided by 4. This is because any exponent \( n \) can be written as \( 4k + r \), where \( r \) could be 0, 1, 2, or 3, aligning it perfectly with the known cycle. This simplification reduces complex calculations dramatically, enabling us to find the equivalent power from the cycle.
- \( i^1 = i \)
- \( i^2 = -1 \)
- \( i^3 = -i \)
- \( i^4 = 1 \)
This knowledge allows us to find the power of any large exponent involving \( i \) by determining its remainder when divided by 4. This is because any exponent \( n \) can be written as \( 4k + r \), where \( r \) could be 0, 1, 2, or 3, aligning it perfectly with the known cycle. This simplification reduces complex calculations dramatically, enabling us to find the equivalent power from the cycle.
Simplification of Algebraic Expressions
Simplifying algebraic expressions with complex numbers involves using the cycles of powers for \( i \). When faced with a large power like \( i^{100} \), rather than computing it step-by-step through repeated multiplication, identify its equivalent from the known cycle.
To simplify \( i^{100} \), you should:
To simplify \( i^{100} \), you should:
- Divide the exponent by 4. For example, \( 100 \div 4 = 25 \), remainder \( 0 \).
- The remainder indicates which part of the \( i \) cycle \( i^{100} \) falls into. Here, a remainder of 0 implies it aligns with \( i^4 = 1 \).
- Thus, \( i^{100} \) simplifies to \( 1 \), making it easy to express in the form \( a + bi \) as \( 1 + 0i \).
Cyclic Patterns in Mathematics
Cyclic patterns form a fundamental concept in mathematics and are prevalent beyond just the imaginary unit. A cyclic pattern repeats a sequence of values after a fixed number of steps—a principle that applies to sequences ranging from the power cycle of the imaginary unit to other repetitive mathematical structures.
With \( i \), the cyclic pattern occurs every four terms, as observed:
With \( i \), the cyclic pattern occurs every four terms, as observed:
- This cyclical nature allows us to use division to identify an exponent's position within a cycle through remainders.
- The concept helps simplify complex calculations into predictable outcomes, as any number raised to a power that is a multiple of the cycle length simplifies to a known value.
Other exercises in this chapter
Problem 81
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