91Ó°ÊÓ

The imaginary unit \(i\) is fundamental to complex numbers. Understanding the powers of \(i\) is crucial for dealing with complex expressions. \(i\) is defined as the square root of \(-1\), so \(i^1 = i\). Through calculation, we find the following:

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

Notice that after the fourth power, the cycle repeats. This cyclical behavior makes it simple to determine higher powers of \(i\) without extensive calculation. For any power \(n\), \(i^n\) will be equivalent to one of these four results. This pattern is recurring every four consecutive powers.

Simplifying expressions involving powers of \(i\) is straightforward once you've understood the cycle of powers. Here’s the process:To find \((-i)^{13}\), first note that multiplying \(-1\) by itself any odd number of times results in \(-1\), while an even number yields \(1\). Additional negative sign in \(-i\) needs to be accounted for when raising to a power. Find \(i^{13}\), knowing that dividing 13 by 4 gives a remainder of 1. Hence, \(i^{13} = i^1 = i\). Thus, for \(-i^{13}\), it initially gives \(-i\). The extra negative sign from \(-i\) simplifies to \(i\) after canceling with another negative, because \(-(-i) = i\). Thus, \((-i)^{13} = i\). The simplification becomes easy with understanding of these core concepts.

The cyclical pattern of \(i\) provides a shortcut for computations involving any power of \(i\). This cycle repeats every four powers. It includes:

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

By using this cyclical nature, instead of calculating large powers directly, - Divide the exponent by 4- Use the remainder to find the equivalent lower power of \(i\)For example, consider the cycle when addressing \(i^{13}\). Since 13 divided by 4 has a remainder of 1, it maps directly onto \(i^1\), which simplifies to \(i\).This cyclical pattern not only saves time but also helps in achieving consistent simplification, making working with powers of \(i\) more manageable.