Problem 83 - A Graphical Approach to College Algebra

Question

Simplify each power of i to \(i, 1,-i,\) or \(-1\). $$\frac{1}{i^{-51}}$$

Step-by-Step Solution

Answer

The simplified result is \(-i\).

Step 1: Understand Powers of i

The imaginary number \(i\) is defined as \(i = \sqrt{-1}\). The powers of \(i\) cycle every four: \(i^1 = i\), \(i^2 = -1\), \(i^3 = -i\), and \(i^4 = 1\). After that, the cycle repeats.

Step 2: Simplify the Negative Exponent

To simplify \(i^{-51}\), first understand that a negative exponent indicates the reciprocal of the positive exponent. So, \(i^{-51} = \frac{1}{i^{51}}\). However, since we have this division involving \(i^{-51}\), simplification can be done by considering \(i^{51}\) directly.

Step 3: Find the Equivalent Positive Power

Since the powers of \(i\) repeat every four, we need to find the remainder when 51 is divided by 4. Divide 51 by 4 to get:\[51 \div 4 = 12 \text{ remainder } 3\]So, \(i^{51}\) is equivalent to \(i^3\).

Step 4: Evaluate i to the Calculated Power

From the cycle of \(i\)'s powers, we know that \(i^3 = -i\). Thus, \(i^{51} = i^3 = -i\).

Step 5: Simplify the Original Expression

Now substitute \(i^{51}\) with \(-i\) in the original expression. Therefore,\[\frac{1}{i^{-51}} = i^{51} = -i\]

Step 6: Verify the Result

Double check the calculation by referring back to the cycle of \(i\): \(i^1 = i\), \(i^2 = -1\), \(i^3 = -i\), \(i^4 = 1\). Since we found \(i^{51} = i^3 = -i\), the simplification is correct.

Key Concepts

Imaginary Unit

In mathematics, the imaginary unit, denoted as \(i\), is a fundamental component of complex numbers. The imaginary unit is defined by the equation \(i = \sqrt{-1}\). This means that \(i\) is a number which, when squared, yields \(-1\). This concept allows mathematicians and scientists to work with equations that would otherwise have no solution within the realm of real numbers.To visualize the impact of the imaginary unit, remember that it offers a way to "step outside" the real number line. While a traditional number you handle in day-to-day life can be found somewhere on the real number line, adding a dimension for the imaginary unit creates a complex plane, making calculations in fields like engineering and physics possible.

• \(i\) is crucial for expressing solutions to polynomial equations that lack real roots.
• Complex numbers, such as \(3 + 4i\), incorporate a real part and an imaginary part.

If you ever need to remember \(i\), think of it as a mathematical tool to understand and solve equations that step beyond typical numerical boundaries.

Negative Exponent

A negative exponent might seem intimidating at first, but it's actually fairly simple once you grasp the concept! When you encounter a negative exponent in mathematics, it signifies the reciprocal of the positive exponent. For example, \(x^{-n} = \frac{1}{x^n}\).This principle applies not only to standard numbers but also to complex numbers and imaginary numbers like \(i\). With \(i^{-51}\), it translates to \(\frac{1}{i^{51}}\). This conversion helps simplify expressions and calculations, making it a powerful tool in algebra and calculus.

• A negative exponent flips a base from the numerator to the denominator.
• This approach is crucial for simplifying many mathematical expressions, ensuring that computations are straightforward and manageable.

The takeaway: Negative exponents are not just about flipping, but they also pave the way to deeper understanding and application of exponential functions across different areas of mathematics.

Exponents Cycle

Understanding the exponents cycle, especially for the imaginary unit \(i\), can greatly simplify calculations involving powers. The powers of \(i\) showcase a repeating pattern every four exponents:

• \(i^1 = i\)
• \(i^2 = -1\)
• \(i^3 = -i\)
• \(i^4 = 1\)

Once you reach the fourth power, the sequence begins again, like clockwork. This predictable cycle allows mathematicians to simplify expressions with higher powers of \(i\) efficiently by using modular arithmetic.For example, to simplify \(i^{51}\), you only need to find 51 mod 4, which leaves a remainder of 3. Thus, \(i^{51}\) behaves the same as \(i^3\), both simplifying to \(-i\). This cycle mechanism is immensely useful:

• It saves countless steps in computation, especially in higher mathematics.
• Recognizing this cycle turns an otherwise complicated problem into an easy task.

The repetitive nature of the exponents cycle is a brilliant trick where simplicity reigns, bringing clarity and elegance to complex number calculations.