Problem 91
Question
Simplify each expression. See Example 11 . $$ i^{27} $$
Step-by-Step Solution
Verified Answer
The simplified form of \( i^{27} \) is \( -i \).
1Step 1: Understanding the Properties of 'i'
The imaginary unit, often denoted as 'i', is defined by the property that \[ i^2 = -1 \].This property creates a cyclic pattern when powers of 'i' are considered. Specifically, the powers of 'i' repeat every four terms:\[ i^1 = i, \; i^2 = -1, \; i^3 = -i, \; i^4 = 1 \]. After this cycle, the pattern repeats.
2Step 2: Finding the Remainder of 27 Divided by 4
To determine which term in the cycle corresponds to \(i^{27}\), we need to find the remainder when 27 is divided by 4 (since the powers of 'i' repeat every 4 terms). Perform the division: \[ 27 \div 4 = 6 \text{ remainder } 3 \]. This means 27 can be expressed as: \[ 27 = 4 \times 6 + 3 \].
Key Concepts
Powers of iCyclic PatternModular Arithmetic
Powers of i
Imaginary numbers often start with understanding the unit imaginary number 'i'. One fundamental property of 'i' is understanding that
- \( i^2 = -1 \)
- \( i^1 = i \)
- \( i^2 = -1 \)
- \( i^3 = -i \)
- \( i^4 = 1 \)
Cyclic Pattern
The powers of 'i' demonstrate a cyclic pattern that repeats every four terms. This is a cycle that every student should memorize as it simplifies complex calculations involving 'i'. Let's break it down:
- The sequence begins with: \( i, -1, -i, 1 \)
- Then, it continues again: \( i^{5}= i, i^{6}= -1 \), and so on.
- Recognizing this four-term cycle is crucial in simplifying expressions efficiently.
Modular Arithmetic
Modular arithmetic plays a vital role when working with powers of 'i'. This framework helps find where a power falls within the established cycle. To determine the equivalent power of a large exponent like \( i^{27} \), modular arithmetic comes into play.
The initial step is performing a division:
This method reliably reduces any power of 'i' into a manageable form by finding its equivalent remainder. Understanding and mastering modular arithmetic helps streamline calculations, making complex number problems less intimidating.
The initial step is performing a division:
- Divide 27 by 4, the cycle length.
- We calculate: \( 27 \div 4 = 6 \) with a remainder of 3.
This method reliably reduces any power of 'i' into a manageable form by finding its equivalent remainder. Understanding and mastering modular arithmetic helps streamline calculations, making complex number problems less intimidating.
Other exercises in this chapter
Problem 91
Simplify each expression, if possible. All variables represent positive real numbers. $$ \sqrt[5]{x^{6} y^{2}}+\sqrt[5]{32 x^{6} y^{2}}+\sqrt[5]{x^{6} y^{2}} $$
View solution Problem 91
Rationalize each denominator. All variables represent positive real numbers. $$ \frac{\sqrt{2}}{\sqrt{5}+3} $$
View solution Problem 91
Find the distance between each pair of points. \((\sqrt{48}, \sqrt{150})\) and \((\sqrt{12}, \sqrt{24})\)
View solution Problem 91
Solve each equation. Write all proposed solutions. Cross out those that are extraneous. $$ \sqrt{-5 x+24}=6-x $$
View solution