Problem 85 - A Graphical Approach to College Algebra

Question

Simplify each power of i to \(i, 1,-i,\) or \(-1\). $$\frac{-1}{-i^{12}}$$

Step-by-Step Solution

Answer

The expression simplifies to 1.

Step 1: Simplify the Denominator Power of i

The problem deals with the expression \(-i^{12}\). Since each power of \(i\) cycles every four terms, first simplify \(i^{12}\).Recall the power cycle of \(i\):- \(i^1 = i\)- \(i^2 = -1\)- \(i^3 = -i\)- \(i^4 = 1\)The cycle repeats every four.Now calculate \(12 \mod 4 = 0\), so \(i^{12} = (i^4)^3 = 1^3 = 1\).

Step 2: Simplify the Entire Expression

Now that we know \(i^{12} = 1\), substitute in the expression:\[\frac{-1}{-i^{12}} = \frac{-1}{-1}\]Simplifying this fraction results in:\[1\]

Key Concepts

powers of i

Complex numbers are often expressed in terms of the imaginary unit, denoted as \(i\), where \(i\) is defined as \(\sqrt{-1}\). The powers of \(i\) exhibit a repeating pattern that is crucial to simplifying expressions involving complex numbers. This pattern occurs every four powers.

• \(i^1\) equals \(i\) because it's simply the unit itself.
• \(i^2\) equals \(-1\), which comes from the property \(i^2 = (\sqrt{-1})^2 = -1\).
• \(i^3\) equals \(-i\), found by multiplying \(i^2\) by \(i\), which gives \(-1 \times i = -i\).
• \(i^4\) equals \(1\), as \(-i \times i = -i^2 = -(-1) = 1\).

After the fourth power, the sequence repeats: \(i^5 = i\), \(i^6 = -1\), and so on. Recognizing this cycle allows for quick simplification of higher powers of \(i\), such as \(i^{12}\), by using modular arithmetic.

simplification

Simplifying complex expressions with powers of \(i\) involves identifying the repeating pattern of powers of \(i\) and applying these when needed. Once you determine the equivalent power from the cycle, you can substitute it back into your expression for easier computation.
In the given problem, you start by simplifying the denominator \(-i^{12}\). Since the powers of \(i\) repeat every four terms, you calculate \(i^{12}\) by understanding that it is the same as \(i^0\) due to the cycle. This allows you to substitute \(i^{12}\) with \(1\), leading to the simplification of \(\frac{-1}{-1}\). Here, recognizing the power pattern allowed us to treat \(i^{12}\) as \(1\), transforming a seemingly complex expression into a simple one.

• Identify the cycle period.
• Use the calculated power to replace high powers of \(i\).
• Substitute and simplify the expression.

modular arithmetic

Modular arithmetic is a useful tool in simplifying powers of \(i\) by determining the remainder when a number is divided by another. It's often referred to as "clock arithmetic" because it cycles back around after reaching a certain point, much like reading a clock.
To figure out which power of \(i\) corresponds to a large exponent, you use modular arithmetic to find \(n \mod 4\). This process identifies the remainder of \(n\) divided by 4, thereby pinpointing how far into the cycle the power falls.

• For \(i^{12}\), calculate \(12 \mod 4 = 0\). This zero remainder indicates that \(i^{12}\) corresponds to \(i^0 = 1\).
• This relationship helps convert any power of \(i\) into one of four possibilities: \(i, -1, -i, \) or \(1\), simplifying the expression easily.
• Modular arithmetic helps tackle complex problems with simplicity, which is especially useful when dealing with large exponents.

Understanding and using modular arithmetic allows for the efficient simplification of complex expressions involving high powers, making math with complex numbers less daunting.