Problem 1 - Precalculus

Question

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. $$\begin{array}{llllll} i^{2} & i^{3} & i^{4} & i^{5} & i^{6} & i^{7} & i^{8} \\ \hline-1 & & & & & \\ \hline \end{array}$$

Step-by-Step Solution

Answer

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

Step 1: 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 \( i \).

Step 2: 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 \( i^3 = -i \).

Step 3: Find i^4

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

Step 4: 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, \( i^5 = i \).

Step 5: Find i^6

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

Step 6: Find i^7

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

Step 7: Find i^8

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

Step 8: Complete The Table

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

Key Concepts

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!