Powers of Iota: Solved Examples - JEE PDF Download

Introduction

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 ± √3i)/2. Hence, the value of iota is helpful in solving square roots with negative values.
Thus, the value of iota is, i = √−1.

Power of i

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. 

Square of Iota 

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.

Square Root of Iota

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 Power of 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