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Chapter 12: Problem 1

In any computation involving complex numbers, express your answer in the form \(a+b i,\) where a and b are real numbers. If \(a\) or \(b,\) or both are zero, then simplify further. Complete the table. $$\begin{array}{llllll} i^{2} & i^{3} & i^{4} & i^{5} & i^{6} & i^{7} & i^{8} \\ \hline-1 & & & & & \\ \hline \end{array}$$

Short Answer

Expert verified
Each power of \(i\) cycles every 4 terms: \( -1, -i, 1, i \).

Step by step solution

01

Understand the Properties of Imaginary Unit i

The imaginary unit \( i \) is defined such that \( i^2 = -1 \). This is a fundamental property used to derive powers of \( i \).
02

Find i^3

We use the property \( i^2 = -1 \) to compute \( i^3 \). Since \( i^3 = i^2 \cdot i = -1 \cdot i = -i \), we have \( i^3 = -i \).
03

Find i^4

We continue with \( i^4 = i^3 \cdot i = (-i) \cdot i = -i^2 = -(-1) = 1 \). Thus, \( i^4 = 1 \).
04

Find i^5 Using Cyclic Pattern

Powers of \( i \) are cyclic every 4 terms as \( i^5 = i^{4+1} = i^4 \cdot i = 1 \cdot i = i \). Hence, \( i^5 = i \).
05

Find i^6

Since \( i^6 = i^{5+1} = i^5 \cdot i = i \cdot i = i^2 = -1 \), we have \( i^6 = -1 \).
06

Find i^7

Similarly, \( i^7 = i^{6+1} = i^6 \cdot i = -1 \cdot i = -i \). Thus, \( i^7 = -i \).
07

Find i^8

Lastly, \( i^8 = i^{7+1} = i^7 \cdot i = (-i) \cdot i = -i^2 = -(-1) = 1 \). Hence, \( i^8 = 1 \).
08

Complete The Table

Using the results from the previous steps, fill in the table:\[\begin{array}{ccccccc}\hline i^2 & i^3 & i^4 & i^5 & i^6 & i^7 & i^8 \\hline -1 & -i & 1 & i & -1 & -i & 1 \\hline\end{array}\]

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Key Concepts

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

Imaginary Unit
In mathematics, when we hear "imaginary unit," we're talking about the symbol \( i \). This unique number was invented to deal with the square roots of negative numbers.
You might wonder why we need it! Well, in the real number system, you cannot take the square root of a negative number. This is where \( i \) comes in handy. It is defined such that \( i^2 = -1 \).
With this definition:
  • \( i \) represents the square root of \(-1\)
  • It helps expand the number system just like how negative numbers expanded from only positive numbers
  • Allows us to compute and simplify expressions involving negative square roots
This foundational property of \( i \) is key when working with complex numbers.
Powers of i
When dealing with the powers of the imaginary unit, \( i \), you'll find a fascinating pattern unravel. The powers of \( i \) do not create entirely new numbers each time. Let's dig into some common powers of \( i \):
  • \( i^1 = i \), which is simply the imaginary unit itself
  • \( i^2 = -1 \), based on the definition
  • \( i^3 = i^2 \cdot i = -1 \cdot i = -i \)
  • \( i^4 = i^3 \cdot i = (-i) \cdot i = -i^2 = 1 \)
From here, you'll see that the pattern repeats every four powers.
When tasked with higher powers of \( i \), determining the remainder of that power divided by 4 will guide you to which of the four outcomes from above will correspond: \( i \), \(-1\), \(-i\), or \(1\).
This cyclical behavior makes calculations much simpler!
Cyclic Pattern
The cyclic pattern of \( i \)'s powers lets us work efficiently with these numbers. Since the sequence of \( i, -1, -i, 1 \) keeps repeating every four terms, it forms a cycle:
  • \( i^1 = i \)
  • \( i^2 = -1 \)
  • \( i^3 = -i \)
  • \( i^4 = 1 \)
  • \( i^5 = i \) -- back to the start of the cycle!
Understanding this cyclic nature eliminates unnecessary calculations.
Instead of manually multiplying \( i \) repeatedly, you can rely on knowing where you are in this four-step cycle.
For example, if you had to find \( i^{100} \), you could simplify it by noting that 100 divided by 4 leaves a remainder of 0, pointing you to \( i^4 = 1 \) as the value for \( i^{100} \).
This kind of pattern recognition is advantageous in simplifying complex calculations!

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Most popular questions from this chapter

Provides an example in which an error in a partial fraction decomposition is not easily detected with a graphical approach. Indeed, this may be an example of a case in which, to check your partial fractions work, it鈥檚 easier to repeat the algebra than to experiment with numerous viewing rectangles. Decide for yourself after completing the problem. There is an error in the following partial fraction decomposition: $$\frac{1}{(x+2)(x+5)(x-14)}=\frac{-1 / 48}{x+2}+\frac{1 / 57}{x+5}+\frac{1 / 305}{x-14}$$ (a) Let \(f\) and \(g\) denote the two functions defined by the expressions on the left side and the right side, respectively, in the above equation. Use a graphing utility to graph \(f\) and \(g,\) first in the standard viewing rectangle and then in the rectangle [-15,15,5] by \([-0.02,0.04,0.02] .\) In this latter rectangle, note that the graphs do appear to be identical. (People using a software graphing application and looking at the curves on a computer monitor may have a little advantage here over those drawing the graphs on a relatively small graphing calculator screen.) (b) Find a viewing rectangle clearly demonstrating that the graphs of \(f\) and \(g\) are not identical. (c) Find the correct partial fraction decomposition, given that the form is $$\frac{1}{(x+2)(x+5)(x-14)}=\frac{A}{x+2}+\frac{B}{x+5}+\frac{C}{x-14}$$

This exercise completes the discussion of improper rational expressions in this section. (a) Use long division to obtain the following result: \(\frac{2 x^{3}+4 x^{2}-15 x-36}{x^{2}-9}=(2 x+4)+\frac{3 x}{x^{2}-9}\) (b) Find constants \(A\) and \(B\) such that \(3 x /\left(x^{2}-9\right)=A /(x-3)+B /(x+3) .\) (According to the text, you should obtain \(A=B=3 / 2 .\) )

The following result is a particular case of a theorem proved by Professor David C. Kurtz in The American Mathematical Monthly [vol. \(99(1992), \text { pp. } 259-263]\) Suppose we have a cubic equation \(a_{3} x^{3}+a_{2} x^{2}+a_{1} x+a_{0}=0\) in which all of the coefficients are positive real numbers. Furthermore, suppose that the following two inequalities hold. $$ a_{1}^{2}>4 a_{0} a_{2} \quad \text { and } \quad a_{2}^{2}>4 a_{1} a_{3} $$ Then the cubic equation has three distinct real roots. (a) Check that these inequalities are valid in the case of the equation \(2 x^{3}+8 x^{2}+7 x+1=0 .\) This implies that the equation has three distinct real roots. Use a graphing utility to verify this and to estimate each root to the nearest one hundredth. (b) Follow part (a) for the equation \(3 x^{3}+40 x^{2}+100 x+30=0\) (c) Use a graphing utility to demonstrate that the graph of \(y=6 x^{3}+15 x^{2}+11 x+2\) has three distinct \(x\) -intercepts. Thus, the equation \(6 x^{3}+15 x^{2}+11 x+2=0\) has three distinct real roots. Now check that the condition \(a_{2}^{2}>4 a_{1} a_{3}\) fails to hold in this case. Explain why this does not contradict the result from Professor Kurtz stated above.

Express the polynomial \(f(x)\) in the form \(a_{n} x^{n}+a_{n-1} x^{n-1}+\cdots+a_{1} x+a_{0}\). (a) Find a third-degree polynomial function that has zeros \(-5,2,\) and 3 and a graph that passes through the point (0,1) (b) Use a graphing utility to check that your answer in part (a) appears to be correct.

\(\frac{x^{3}+4 x^{2}+16 x}{x^{4}+16}\) Hint: To factor the denominator, add and subtract the term \(16 x^{2}\)
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