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Chapter 3: Problem 76

Simplify each expression to \(i, 1,-i,\) or \(-1\) $$i^{19}$$

Short Answer

Expert verified
The simplified form of \(i^{19}\) is \(-i\).

Step by step solution

01

Understanding the Power Cycle of i

The imaginary unit, represented by \(i\), follows a specific cycle when raised to successive powers. This cycle is: 1. \(i^1 = i\)2. \(i^2 = -1\)3. \(i^3 = -i\)4. \(i^4 = 1\) This pattern repeats every four powers.
02

Divide the Exponent by the Cycle Length

Since \(i\) has a cycle of 4, divide the exponent 19 by 4 to find the remainder. This can be done using integer division. Calculating: \[ 19 \div 4 = 4 ext{ remainder } 3 \]. The remainder is 3.
03

Use the Remainder to Determine the Result

The remainder of 3 from the previous step tells us that \(i^{19}\) has the same value as \(i^3\). From the power cycle established earlier, we know that: \[ i^3 = -i \].

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Imaginary Unit
The imaginary unit, denoted as \(i\), is fundamental in working with complex numbers. It represents the square root of -1. Typically, we learn in mathematics that negative numbers don't have real square roots. However, \(i\) allows us to handle these roots by providing a way to work with them in calculations.
For clarity, remember:
  • \(i^2 = -1\) - The defining property of the imaginary unit.
  • \(i\) itself is a representation, not a real number, but something mathematically accurate to facilitate calculations.
Understanding \(i\) is crucial when dealing with complex numbers because it exists outside of the regular number line, adding dimensions to our mathematical operations.
Powers of i
Powers of \(i\) form a recurring cycle, so it's essential to recognize the pattern:
  • \(i^1 = i\)
  • \(i^2 = -1\)
  • \(i^3 = -i\)
  • \(i^4 = 1\)
This cycle repeats every four numbers. The straightforward nature of this cycle makes powers of \(i\) simple to handle in calculations. When faced with a large exponent, like \(i^{19}\), you can simplify the problem by using the modulo operation to reduce the exponent. For instance, \(19 \div 4 = 4\) remainder \(3\), meaning \(i^{19} = i^3\). Knowing the cycle helps us easily determine that \(i^{19}\) equals \(-i\).
The cycle helps in quickly finding equivalent expressions for the powers of \(i\), simplifying many equations that include complex numbers.
Simplification of Expressions
Simplifying expressions involving complex numbers is a key skill, especially when dealing with imaginary units and their powers. The main goal is to reduce expressions into their simplest form, frequently an integer or one of \(i, 1, -i,\) or \(-1\).
The procedure you need to follow generally involves:
  • Identifying the power of \(i\).
  • Using the known cycle of \(i\) powers to find an equivalent expression within the first four powers.
For example, in the problem \(i^{19}\), by knowing the cycle length is 4, you determine \(19 \mod 4 = 3\), directing you to \(i^3\). From there, using the pattern, you know \(i^{19} = -i\).
This method allows you to handle even the most complex-looking expressions involving powers of \(i\) without much effort, as long as you remember the cycle and use modulus to simplify.

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Most popular questions from this chapter

Solve each problem. An athlete's heart rate \(R\) in beats per minute after \(x\) minutes is given by $$R(x)=2(x-4)^{2}+90.$$ where \(0 \leq x \leq 8\) (a) Describe the heart rate during this period of time. (b) Determine the minimum heart rate during this 8 -minute period.

Solve each equation. For equations with real solutions, support your answers graphically. $$(x+5)(x-6)=(2 x-1)(x-4)$$

Solve each formula for the indicated variable. Leave \(\pm\) in answers when applicable. Assume that no denominators are 0 $$V=\frac{4}{3} \pi r^{3} \quad \text { for } r$$

For each pair of numbers, find the values of \(a, b,\) and \(c\) for which the quadratic equation ax \(^{2}+b x+c=0\) has the given numbers as solutions. Answers may vary. (Hint: Use the zero-product property in reverses $$2-\sqrt{5}, 2+\sqrt{5}$$

Solve each quadratic equation by completing the square. $$2 x^{2}+6 x-3=0$$
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