Power of Iota

The nature of a quadratic equation’s roots depends entirely on the value of its discriminant. When the discriminant is negative, the equation has complex roots, which involve the imaginary unit iota (i).

The imaginary unit, denoted as i, is defined as: i = √-1

For example, consider the quadratic equation x² + x + 1 = 0.
Using the Quadratic Formula:
Here, the discriminant (b² − 4ac) is negative, leading to a square root of a negative number. In such cases, we use iota to express the solution, as: i = √−1​
Thus, the concept of i allows us to work with numbers beyond the real number system, forming the foundation of complex numbers.

Fundamental Powers of i

1. i¹ = i
2. i² = −1
3. i³ = −i
4. i⁴ = 1

Cyclic Nature of i

Observing the powers above, we notice that they repeat every four steps. This means that for any integer exponent n:

iⁿ = i^{(n mod  4)}
where n mod gives the remainder when n is divided by 4.

Generalization of Higher Powers of i

For any integer n:

• If n ≡ 0 mod 4, then iⁿ = 1
• If n ≡ 1 mod 4, then iⁿ = i
• If n ≡ 2 mod 4, then iⁿ = −1
• If n ≡ 3 mod 4, then iⁿ = −i

Examples

1. i¹⁰
   • Find 10 mod  4 = 2
   • Since i² = −1 , we get i¹⁰ = −1
2. i²³
   • Find 23 mod  4 = 3
   • Since i³ = −i
3. i¹⁰⁰
   • Find 100 mod  4 = 0
   • Since i⁴ = 1, we get i¹⁰⁰ = 1

Note: The results will always be one of 1, i, −1, or −i.

Important Formulas Power of Iota

Different powers of Iota are shown below, this table shows the calculations and results for the powers of the imaginary unit i.

This signifies that i repeats its values after every 4th power. We can generalize this fact to represent this pattern (where n is any integer), as,

• i⁴ⁿ = 1
• i⁴ⁿ⁺¹ = i
• i⁴ⁿ⁺² = -1
• i⁴ⁿ⁺³ = -i

Solved Examples of Power of Iota

Example 1: Simplify i⁵¹⁷

Solution:

After dividing power of 'i' by 4 remainder is 1
= i⁵¹⁷
= i¹
= i

Example 2: Solve i²⁰⁹⁵

Solution:

After dividing power of 'i' by 4 remainder is 3
= i²⁰⁹⁵
= i²
= -i

Example 3: Solve i²³⁴⁶

Solution:

After dividing power of 'i' by 4 remainder is 0
= i²³⁴⁵⁶
= i⁰
= 1

Example 4: Solve i³²⁴⁷⁷⁰

Solution:

After dividing power of 'i' by 4 remainder is 2
i³²⁴⁷⁷⁰
= i²
= -1

Example 5: Find the value of i^-3927

Solution:

i^-3927 = 1/i³⁹²⁷ {∵ a^-m = 1/a^m}
= 1/i³

After dividing power of 'i' by 4 remainder is 3

= 1/-i {∵ i³ = -i}
= -(-i) {∵ 1/i = -i}
= i

Practice Problems on Power of Iota

Question 1: Find the value of i^-432

Question 2: Find the value of i^-12

Question 3: Find the value of i²⁶

Question 4: Find the value of i^-512

Question 5: Find the value of i¹⁰²⁴

Question 6: Find the value of i^-2048

Question 7: Find the value of i¹²¹

Question 8: Find the value of i²³³

Related Article:

• Complex Number
• Complex Number Power Formula
• Value of i in Complex Number
• Imaginary Numbers