Problem 108
Question
Suppose that your friend describes a method of simplifying a positive power of \(i .\) "Just divide the exponent by 4 and then look at the remainder. Then, refer to the short table of powers of \(i\) in this section. The given power of \(i\) is equal to \(i\) to the power indicated by the remainder. And if the remainder is \(0,\) the result is \(i^{0}=1 .^{\prime \prime}\) Explain why this method works.
Step-by-Step Solution
Verified Answer
The method works because powers of \(i\) cycle every four terms.
1Step 1: Understanding the Exponents of i
The imaginary unit, denoted as \(i\), has certain properties when raised to integer powers. The key is knowing that \(i^n\) is cyclical with a period of 4: \(i^1 = i\), \(i^2 = -1\), \(i^3 = -i\), \(i^4 = 1\), and then it repeats. This cyclical behavior is crucial in simplifying powers of \(i\).
2Step 2: Divide the Exponent by 4
When dividing the exponent of \(i\) by 4, the operation is essentially reducing the problem to one of four cases: remainder 0, 1, 2, or 3. This correlates with \(i^0\), \(i^1\), \(i^2\), and \(i^3\) respectively.
3Step 3: What the Remainder Indicates
The remainder determines which point in the 4-step cycle the exponent falls on: - Remainder 0 corresponds to \(i^4 = 1\).- Remainder 1 corresponds to \(i^1 = i\).- Remainder 2 corresponds to \(i^2 = -1\).- Remainder 3 corresponds to \(i^3 = -i\).
4Step 4: Conclusion of Why This Method Works
The method works because it leverages the periodic nature (modulo 4) of powers of \(i\). By reducing any power to these four basic outcomes, we simplify calculations and avoid exponentiation directly.
Key Concepts
Imaginary UnitPowers of iCyclic Patterns
Imaginary Unit
The concept of the imaginary unit is fundamental in extending our understanding of numbers beyond the real number line. The imaginary unit is denoted as \(i\), and it is defined by the property \(i^2 = -1\). This definition may seem strange at first, especially because no real number squared gives a negative result. However, the introduction of \(i\) allows us to solve equations that have no real solutions, like \(x^2 + 1 = 0\). This enriches the set of numbers mathematicians work with, leading to complex numbers. Complex numbers are expressions of the form \(a + bi\), where \(a\) and \(b\) are real numbers, and \(b\), includes a part that is multiplied by \(i\). This opens up a universe that leverages the imaginary unit to find solutions that reach beyond the limits of real numbers.
Powers of i
When we raise the imaginary unit \(i\) to various powers, an interesting pattern emerges, which is both simple and highly practical for solving complex problems. Here's what happens:
This cycle of four allows us to always express any power of \(i\) in one of four basic forms: \(i\), \(-1\), \(-i\), or \(1\). This repetition makes dealing with powers of \(i\) much more manageable and predictable. Whenever faced with a large exponent, break it down by dividing the exponent by 4, using the remainder to determine which of these four values it corresponds to.
- \(i^1 = i\)
- \(i^2 = -1\)
- \(i^3 = -i\)
- \(i^4 = 1\)
This cycle of four allows us to always express any power of \(i\) in one of four basic forms: \(i\), \(-1\), \(-i\), or \(1\). This repetition makes dealing with powers of \(i\) much more manageable and predictable. Whenever faced with a large exponent, break it down by dividing the exponent by 4, using the remainder to determine which of these four values it corresponds to.
Cyclic Patterns
Cyclic patterns in mathematics refer to sequences that repeat themselves periodically after a certain number of steps. Powers of \(i\) is a perfect example. The pattern \(i, -1, -i, 1\) repeats every four steps, creating a cycle. This phenomenon, known as a 'cyclic pattern,' is observed in various mathematical contexts, such as trigonometry and number theory.
- The cyclic pattern of powers of \(i\) consists of four stages.
- At the end of each cycle, the sequence begins anew.
- This concept simplifies calculations, as you only need to focus on the remainder when the exponent is divided by 4.
Other exercises in this chapter
Problem 107
Explain why the method of dividing complex numbers (that is, multiplying both the numerator and the denominator by the conjugate of the denominator) works. What
View solution Problem 108
Solve each inequality analytically. Support your answers graphically. Give exact values for endpoints. (a) \(3 x^{2}+13 x \leq-10\) (b) \(3 x^{2}+13 x>-10\)
View solution Problem 109
Solve each inequality analytically. Support your answers graphically. Give exact values for endpoints. (a) \(-x^{2}-x \leq 0\) (b) \(-x^{2}-x>0\)
View solution Problem 110
Solve each inequality analytically. Support your answers graphically. Give exact values for endpoints. (a) \(-x^{2}+2 x \leq 0\) (b) \(-x^{2}+2 x>0\)
View solution