Understanding the Powers of the Imaginary Unit 'i'

The question asks us to find the value of the sum \( i^{1000} + i^{1001} + i^{1002} + i^{1003} \), where \( i = \sqrt{-1} \). To solve this, we need to understand the pattern of the powers of the imaginary unit \( i \).

The powers of \( i \) follow a cyclical pattern that repeats every four terms:

• \( i^1 = i \)
• \( i^2 = -1 \)
• \( i^3 = i^2 \times i = -1 \times i = -i \)
• \( i^4 = i^2 \times i^2 = -1 \times -1 = 1 \)
• \( i^5 = i^4 \times i = 1 \times i = i \)
• \( i^6 = i^4 \times i^2 = 1 \times -1 = -1 \)
• And so on...

This means that for any integer \( n \), the value of \( i^n \) depends on the remainder when \( n \) is divided by 4. We can write this as \( i^n = i^{n \pmod 4} \), where if the remainder is 0, the power is \( i^4 = 1 \).

Calculating Each Power of i

Let's calculate each term in the sum \( i^{1000} + i^{1001} + i^{1002} + i^{1003} \) using the cyclical property of the powers of \( i \).

• For \( i^{1000} \): We divide 1000 by 4. \( 1000 \div 4 = 250 \) with a remainder of 0. Since the remainder is 0, \( i^{1000} = i^4 = 1 \).
• For \( i^{1001} \): We divide 1001 by 4. \( 1001 = 4 \times 250 + 1 \). The remainder is 1. So, \( i^{1001} = i^1 = i \).
• For \( i^{1002} \): We divide 1002 by 4. \( 1002 = 4 \times 250 + 2 \). The remainder is 2. So, \( i^{1002} = i^2 = -1 \).
• For \( i^{1003} \): We divide 1003 by 4. \( 1003 = 4 \times 250 + 3 \). The remainder is 3. So, \( i^{1003} = i^3 = -i \).

Summing the Calculated Powers of i

Now we substitute the calculated values back into the original expression:

\( i^{1000} + i^{1001} + i^{1002} + i^{1003} = (i^{1000}) + (i^{1001}) + (i^{1002}) + (i^{1003}) \)

\( = (1) + (i) + (-1) + (-i) \)

\( = 1 + i - 1 - i \)

Now, we group the real and imaginary terms:

\( = (1 - 1) + (i - i) \)

\( = 0 + 0 \)

\( = 0 \)

Thus, the sum \( i^{1000} + i^{1001} + i^{1002} + i^{1003} \) is equal to 0.

Generalization: Sum of Consecutive Powers of i

It's worth noting a general property: the sum of any four consecutive integer powers of \( i \) is always 0.

Let's consider the sum \( i^n + i^{n+1} + i^{n+2} + i^{n+3} \).

We can factor out \( i^n \):

\( i^n + i^{n+1} + i^{n+2} + i^{n+3} = i^n (1 + i^1 + i^2 + i^3) \)

We know that \( 1 + i^1 + i^2 + i^3 = 1 + i + (-1) + (-i) = 1 + i - 1 - i = 0 \).

So, \( i^n (1 + i + i^2 + i^3) = i^n (0) = 0 \).

Since 1000, 1001, 1002, and 1003 are four consecutive integers, their powers of \( i \) will sum to 0.

Revision Table: Key Concepts

Review the essential concepts related to powers of the imaginary unit \( i \).

Additional Information: Complex Numbers

The imaginary unit \( i \) is a fundamental part of complex numbers. A complex number is generally written in the form \( a + bi \), where \( a \) and \( b \) are real numbers, and \( i \) is the imaginary unit (\( \sqrt{-1} \)).

• The term \( a \) is called the real part.
• The term \( bi \) is called the imaginary part, and \( b \) is the imaginary coefficient.
• Complex numbers extend the concept of real numbers and are used in various fields like physics, engineering, and mathematics to solve problems that cannot be solved using real numbers alone (like finding the square root of a negative number).
• Operations like addition, subtraction, multiplication, and division can be performed on complex numbers. Understanding the powers of \( i \) is crucial for these operations, especially multiplication and division.