Problem 79 - Algebra A Combined Function

Question

Find each power of \(i\). $$ (2 i)^{6} $$

Step-by-Step Solution

Answer

The result of \((2i)^6\) is \(-64\).

Step 1: Understand Powers of i

The imaginary unit \(i\) is defined such that \(i^2 = -1\). Powers of \(i\) cycle every four: \(i^1 = i\), \(i^2 = -1\), \(i^3 = -i\), \(i^4 = 1\), \(i^5 = i\), and so on. Therefore, for any integer \(n\), the value of \(i^n\) can be determined by \(n \mod 4\).

Step 2: Rewrite the Complex Base

The expression \((2i)^6\) can be expanded using properties of exponents. It can be rewritten as \(2^6 \cdot i^6\).

Step 3: Calculate Powers of the Coefficient

Calculate \(2^6\), which is the coefficient part of the expression. \(2^6 = 64\).

Step 4: Calculate the Relevant Power of i

Since powers of \(i\) repeat every 4, calculate \(i^6\). Find \(6 \mod 4 = 2\). So, \(i^6 = i^2 = -1\).

Step 5: Multiply Results

Combine the results from Steps 3 and 4: \(64 \cdot (-1) = -64\).

Step 6: Conclude the Result

The power \((2i)^6\) simplifies to \(-64\).

Key Concepts

Imaginary Unit

The imaginary unit, denoted as \(i\), is a fundamental concept in the study of complex numbers. It is defined by the property that \(i^2 = -1\). This is a crucial departure from the real numbers, where the square of any real number is positive. The introduction of \(i\) allows us to work with numbers that are not on the traditional number line, expanding our ability to solve equations that do not have real solutions, such as \(x^2 + 1 = 0\).

In everyday mathematical practice, \(i\) functions almost like a variable, but with its unique property that \(i^2 = -1\). This property is essential for tackling problems involving square roots of negative numbers, enabling a whole new dimension of problem-solving.

Understanding \(i\) is a stepping stone towards grasping more complex concepts, such as complex numbers and their operations.

Powers of Imaginary Numbers

When working with powers of the imaginary unit \(i\), it's important to recognize the cyclic pattern that emerges. This cyclic pattern repeats every four powers:

• \(i^1 = i\)
• \(i^2 = -1\)
• \(i^3 = -i\)
• \(i^4 = 1\)
• \(i^5 = i\), continuing the cycle

Recognizing this pattern allows for efficient computation of any power of \(i\). To find \(i^n\) for a given \(n\), you can simply calculate \(n \mod 4\) and use the remainder to determine the value of \(i^n\):

• If the remainder is 0, \(i^n = 1\)
• If the remainder is 1, \(i^n = i\)
• If the remainder is 2, \(i^n = -1\)
• If the remainder is 3, \(i^n = -i\)

This cyclical behavior provides a shortcut, making it easy to calculate large powers of \(i\) without needing extensive computation.

Exponentiation of Complex Numbers

Exponentiation of complex numbers involves raising a complex number, such as \((2i)\), to a power, as seen in expressions like \((2i)^6\). These calculations combine properties of exponents and the unique behavior of the imaginary unit \(i\).

The initial step involves separating the expression into a real component and an imaginary component using power rules: \((2i)^6 = 2^6 \cdot i^6\). Calculating \(2^6\) gives us 64, which represents the real number part. The second part requires evaluating \(i^6\), which can be determined through the cyclic nature of powers of \(i\). Since \(6 \mod 4 = 2\), we know \(i^6 = i^2 = -1\).

Finally, combining the results, we compute \(64 \cdot (-1) = -64\). This result shows how complex number exponentiation leverages both numerical calculations and the patterns inherent to the imaginary unit. Understanding this concept is vital for advanced mathematical computations, especially in fields like engineering and physics.