Powers of iota

Iota is an imaginary unit number that is denoted by i and the value of iota is √-1 i.e., i = √−1. While solving quadratic equations, you might have come across situations where the discriminant is negative. For example, consider the quadratic equation x² + x + 1 = 0. If we use the quadratic formula to solve this, we get the discriminant (the part inside the square root) as a negative value.

In such cases, we write √−3 as √−3 = √−1 × √3. This would give the solution of the above quadratic equation to be: x = (−1 ± √3\(i\))/2. Hence, the value of iota is helpful in solving square roots with negative values.

Thus, the value of iota is, i = √−1.

The powers of i, i repeat in a certain pattern in a cycle. Let us start with calculating the value of powers of i for general cases and try to figure out the pattern. 

We know that the value of iota, i is defined as, i = √−1. If we square both sides of the above equation, we get: i² = -1 i.e., the value of the square of iota is -1. Therefore, the square of iota is,  i² = −1.

Iota has two square roots, just like all non-zero complex numbers. The value of the square root of iota, given as √i, can be calculated using De Moivre's Theorem. 

We know that, i = cos(π/2) + isin(π/2)

= cos(π/2 + 2nπ) + isin(π/2 + 2nπ), n = 0, 1

= cos[(π + 4nπ)/2] + i sin[(π + 4nπ)/2]

Here, we took n = 0, 1 as we need 2 solutions. But we have to find √i = (i)^1/2. Let us raise the exponent to 1/2 on both sides. So we get: √i = [cos{(π + 4nπ)/2} + isin{(π + 4nπ)/2}]^1/2 = cos[(π + 4nπ)/4] + isin[(π + 4nπ)/4], n = 0, 1

• When n = 0, √\(i\) = cos(π/4) + isin(π/4) = √2/2 + i√2/2  
• When n = 1, √\(i\) = cos(5π/4) + isin(5π/4) = −√2/2 − i√2/2

√i = √2/2 + i√2/2 = −√2/2 − i√2/2

Let us see how to calculate some other powers of i.

• i³ = i × i² = i × −1 = −i
• i⁴ = i² × i² = −1×−1 = 1
• i⁵ = i × i⁴ = i × 1 = i
• i⁶ = i × i⁵ = i × i = i² = −1
• i⁷ = i × i⁶ = i × −1 = -i
• i⁸ = ((i)²)⁴ = (−1)⁴ =1
• i⁹ = i × i⁸ = i × 1 = i
• i¹⁰ = i × i⁹ = i × i = i² = −1

From the above calculations, we can observe that the values of iota repeat in certain pattern. The following figure represents the values for various powers of i in the form of a continuous circle.

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

Higher powers of iota can be calculated by decomposing the higher exponents \(i\) into smaller ones and thus evaluating the expression. Finding value if the power of i is a larger number using the previous procedure, will take quite some time and effort. If we observe all the powers of i and the pattern in which it repeats its values in the above equations, we can calculate the value of iota for higher powers as given below, 

• Step 1: Divide the given power by 4.
• Step 2: Note the remainder for the division in Step 1, and use it as the new exponent/power of i.
• Step 3: Calculate the value of iota for this new exponent/power using the previously known values, i = √−1; i² = -1 and i³ = -i. 

Example: Find the value of i²⁰²⁹⁶.

• We first divide 20296 by 4 and find the remainder.
• The remainder is 0 (by divisibility rules, we can just divide the number formed by the last two digits which is 96 in this case, to find the remainder).
• Thus, using the above rules, i²⁰²⁹⁶ = i⁰ = 1
• Therefore, i²⁰²⁹⁶ = 1

We just have to remember that i² = -1 and i³ = -i. We will find some other higher powers of i using these and the above rules.

Value of Iota for Negative Power

The value of iota for negative power can be calculated following few steps. We first convert it into a positive exponent using the negative exponent law and then we apply the rule: 1/i = -i. This is because:1/i = 1/i • i/i = i/i² = i/(-1) = -i 

Example: Find the value of i^-3927

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

∵ Remainder of 3927 divided by 4 is 3

= 1/-i
∵ i³ = -i
= --i
∵ 1/i = -i
= i

Here is the "Powers of Iota Calculator".You can enter any exponent (positive or negative) of i and see the result in a step-by-step manner.

Important Notes

• The value of iota is i = √−1
• The value of the square of iota is, i² = −1
• The value of the square root of iota is, 
  √i = √2/2 + i√2/2

Tips and Tricks

• To find any power of iota, say iⁿ, just divide n by 4 and find the remainder, r. Then just apply iⁿ = i^r. Here, you just need to remember two things i² = −1 and i³ = −i
• To calculate the negative powers of iota, we use the rule 1/i = −i

What is Iota in Mathematics?

Iota is an imaginary unit number to express complex numbers, where i is defined as imaginary or unit imaginary. Basically, the value of the imaginary unit number, i is generated, when there is a negative number inside the square root. Thus, the value of i is given as √-1. 

What Is the Power Of i?

The power of i is the higher values of i such as i² = -1, i³  = -i, i⁴ = 1. All of these powers of i can be generalized as i⁴ⁿ = 1, i^{4n + 1} = i, i^{4n + 2} = -1, i^{4n + 3} = -i.

What is the Value of i in Mathematics?

The value of the imaginary unit number, i is generated when there is a negative number inside the square root. The value of i in Mathematics is √−1.

What is Iota Cube?

Iota cube is given as i³, which can written as i³ = i²⋅i = (−1)⋅i = −i. Thus, the iota cube is −i.

Who Discovered Iota in Mathematics?

The concept of imaginary numbers in Mathematics was first introduced by mathematician “Euler”. He introduced i (read as 'iota') to represent √-1. Also, he defined i² = -1.

What is the Symbol of Iota?

Iota is an imaginary unit number, denoted using the symbol i.

What is the Power of i to 34?

The value of i power 34 can be calculated as, i³⁴ = ((i)⁴)⁸ • (i)² = 1 × (-1) = -1. Therefore, value of (i)³⁴ = -1.

What is Square Root of Iota?

The value of the square root of iota is, √i = √2/2 + i√2/2.