If i is the imaginary unit with i² = -1, which expression is equivalent to i²³³?

Introduction / Context:
This question involves complex numbers and powers of the imaginary unit i. The imaginary unit is defined by the property i² = -1. Powers of i follow a simple repeating pattern with period 4, which makes evaluating large exponents like i²³³ straightforward once you know the cycle. Such questions commonly appear in algebra sections to test familiarity with complex number basics.

Given Data / Assumptions:

• i is the imaginary unit, so i² = -1.
• We must compute i²³³.
• Arithmetic is in the usual complex number system.

Concept / Approach:
Powers of i repeat in a cycle of length 4: i¹ = i, i² = -1, i³ = -i, i⁴ = 1, and then i⁵ = i again, and so on. Therefore, to evaluate i raised to a large power, we reduce the exponent modulo 4 and identify which one of these four values corresponds to the remainder. The exponent 233 can be divided by 4 to find this remainder.

Step-by-Step Solution:
List the basic cycle: i¹ = i, i² = -1, i³ = -i, i⁴ = 1. The cycle repeats every 4 powers. Compute 233 divided by 4: 4 * 58 = 232 with remainder 1. Thus 233 = 4 * 58 + 1. So i²³³ = i^(4 * 58 + 1) = (i⁴)⁵⁸ * i¹. Since i⁴ = 1, (i⁴)⁵⁸ = 1⁵⁸ = 1. Therefore i²³³ = 1 * i = i.

Verification / Alternative check:
We can test the method on smaller exponents. For example, i⁵ = i^(4 + 1) = i⁴ * i = 1 * i = i, which matches the cycle. For i⁹, we get i^(8 + 1) = (i⁴)² * i = 1 * i = i again. Each time, reducing the exponent modulo 4 and using the remainder as an index in the cycle works. Since 233 leaves a remainder of 1 when divided by 4, the correct value must be the same as i¹, which is i.

Why Other Options Are Wrong:

• 1: This would correspond to exponents that are multiples of 4, such as i⁴, i⁸, i¹² and so on.
• -1: This value appears when the exponent is congruent to 2 modulo 4, for example i², i⁶, i¹⁰.
• -i: This occurs when the exponent is congruent to 3 modulo 4, such as i³ or i⁷.
• 0: No non zero power of i equals zero.

Common Pitfalls:
Students sometimes try to expand i²³³ step by step, which is unnecessary and impossible to manage. Others misremember the cycle or confuse it with other periodic patterns. A reliable approach is always to reduce the exponent modulo 4 because i⁴ = 1 resets the pattern. Writing down the four term cycle and matching the remainder directly to one of i, -1, -i or 1 keeps the process clear and error free.

Final Answer:
The value of i²³³ is i.