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Chapter 2: Problem 45

In Exercises 43 to 50 , evaluate the power of \(i .\) $$-i^{40}$$

Short Answer

Expert verified
The evaluation of \(-i^{40}\) is -1

Step by step solution

01

Determine the modulus

Determine the modulus of the exponent by 4. Calculate \(40 \mod 4\). This gives 0.
02

Evaluate the power of \(i\)

Use the result from Step 1 to evaluate the power of \(i\). If the Modulus is 0, which means the power of \(i\) is a multiple of 4, the result is 1. So, \(i^{40} = 1\)
03

Solution

Substitute the value of \(i^{40}\) into the given expression \(-i^{40} \). Therefore, the answer will be \(-1\*1 = -1\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

powers of i
Understanding the powers of the imaginary unit, \( i \), is crucial to grasping complex numbers. The unit \( i \) is defined as the square root of -1, or \( i^2 = -1 \). From this basic definition, the powers of \( i \) repeat in a predictable cycle:
  • \( i^1 = i \)
  • \( i^2 = -1 \)
  • \( i^3 = -i \)
  • \( i^4 = 1 \)
  • \( i^5 = i \) again, and so forth repeating every four powers.
Whenever you are asked to calculate a higher power of \( i \), you only need to determine its position within this cycle. For example, in the problem \(-i^{40}\), you find that 40 divided by 4 gives a remainder of 0, meaning \( i^{40} \) falls at \( i^4 \), which equals 1.
modulus operation
The modulus operation helps in finding the remainder when a number is divided by another. It is essential in simplifying the powers of \( i \) due to their cyclic nature. For a given integer exponent \( n \), you find the remainder when \( n \) is divided by 4. This is denoted as \( n \mod 4 \).
This means if you have to evaluate \( i^n \), you calculate \( n \mod 4 \) to find which step in the cycle (from \( i^1 \) to \( i^4 \)) corresponds to \( i^n \).
  • If the remainder is 0, \( i^n = i^4 \), and the value is 1.
  • If the remainder is 1, \( i^n = i \).
  • If the remainder is 2, \( i^n = -1 \).
  • If the remainder is 3, \( i^n = -i \).
In the case of \( i^{40} \), the result was 0, indicating \( i^{40} \) simplifies to 1. Hence, \(-i^{40} = -1 \times 1 = -1\). This technique saves a lot of time and effort when dealing with significantly large exponents.
value of imaginary unit i
The imaginary unit, denoted as \( i \), is a fundamental building block in complex numbers. It makes it possible to handle the square roots of negative numbers which are not defined in the set of real numbers.
  • \( i \) is defined as \( \sqrt{-1} \), thus \( i^2 = -1 \).
  • When raised to increasing powers, \( i \)'s values rotate in a predictable sequence: \( i, -1, -i, 1 \).
  • This cyclical pattern is particularly useful as it simplifies complexities of multiplication and division involving \( i \).

Understanding \( i \) gives you insight into how complex numbers behave and how they can be manipulated, making it easier to solve various mathematical problems. In our problem, knowing \( i^{40} = 1 \) helps solve \(-i^{40} = -1\), showcasing the practicality of \( i \)'s properties in resolving such computations.

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