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In mathematics, complex numbers are a natural extension of real numbers and they provide a richer arithmetic ecosystem. They are particularly useful in scenarios that involve square roots of negative numbers and are commonly used in fields such as engineering, physics, and applied mathematics. The standard form of a complex number is denoted by \(a + bi\), where \(a\) and \(b\) are real numbers, while \(i\) is the imaginary unit with the property that \(i^2 = -1\).

When simplifying complex numbers, the goal is often to represent them in standard form to make them easier to interpret and use in further calculations. For instance, the complex number \(\frac{1}{(2i)^3}\) after simplification results in \(-\frac{1}{8}i\), which fits the standard form with \(a = 0\) and \(b = -\frac{1}{8}\). Representing complex numbers in this format is crucial as it allows us to visualize and compare their real and imaginary components directly.

Working with exponential expressions in complex numbers follows rules similar to those applied to real numbers. Specifically, when raising a complex number to an exponent \(n\), we multiply it by itself \(n\) times. For example, to calculate \((2i)^3\), we would perform the multiplication \(2i \times 2i \times 2i\).

However, one needs to keep in mind that powers of \(i\) cycle through a pattern: \(i^1 = i\), \(i^2 = -1\), \(i^3 = i^2 \times i = -1 \times i = -i\), and \(i^4 = 1\). This realization simplifies computations substantially. Therefore, in our exercise, when we simplify \((2i)^3\), we get \(-8i\) because we apply the exponent rule and the cyclical nature of \(i\)'s powers, a concept that's vital for students to recognize.

The concept of multiplying by the conjugate is particularly useful when dealing with complex numbers in fractional forms. The conjugate of a complex number \(a + bi\) is \(a - bi\). When a complex number is multiplied by its conjugate, the result is a real number. This is because the product of \((a + bi)\) and \((a - bi)\) is \(a^2 - (bi)^2\), which simplifies to \(a^2 + b^2\), since \(i^2 = -1\).

Applying this to our problem, to eliminate the imaginary number from the denominator, we multiplied \(-\frac{1}{8i}\) by the conjugate of the denominator, which in this case is also \(-8i\). The multiplication of \(-8i\) by its conjugate yields a real number, specifically \(64\), allowing us to simplify further and obtain the final solution \(-\frac{1}{8}i\). This technique is invaluable as it provides a method to rationalize denominators and simplify complex fractions.