Last updated on July 22nd, 2025
Iota, represented as i, is the imaginary unit with a value of √–1. We use it to calculate the square roots of negative numbers. In this topic, we will discuss the powers of iota and their significance in detail.
In math, Iota is an imaginary unit number (i) with a value of √–1. It can be used to solve equations involving the square root of negative numbers.
For example: √–5 can be expressed as √–1 × √5 = i √5.
Square of Iota:
Since the value of Iota is i = √–1, squaring both sides of the equation gives i2 = –1.
Square Root of Iota
Like every non-zero complex number, iota has two square roots. We use De Moivre’s Theorem to determine the value of the square root of Iota, which we represent as √i.
Since i = cos (π/2) + i sin (π/2)
= cos(π/2 + 2nπ) + i sin(π/2 + 2nπ), n = 0, 1
= cos[(π + 4nπ)/2] + i sin[(π + 4nπ)/2]
The square root of Iota has two solutions:
We use these two values to indicate the square roots of i in complex form.
i1 = i
i2 = –1
i3 = i × i2 = i × –1 = –i
i4 = i2 × i2 = –1 × –1 = 1
The powers of i follow a recurring pattern of 4:
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
To calculate higher powers of Iota, we divide the exponent by 4, and find the remainder with the help of the cyclic pattern. Calculating higher powers using direct multiplication method can be difficult, so we use the following steps or the cyclic pattern:
To calculate the value of iota for negative powers, we follow the steps mentioned below:
The important powers of Iota are shown in the table below:
Since the pattern repeats indefinitely as seen in the table above, we generalize it to denote any number n as:
i4n = 1
i4n + 1 = i
i4n + 2 = -1
i4n + 3= -i
Iota plays a vital role in various real-life situations. Understanding the significance of the powers of iota helps students solve complex problems. Here are a few examples of the applications of the powers of iota:
To solve the complex equations by simplifying them, we use the powers of Iota. But also, students tend to make errors when working with the powers of Iota. Here are a few common errors along with tips to avoid them:
Simplify i70
i70 = –1
We first determine the remainder by dividing 70 by 4:
70 ÷ 4 =17 remainder 2
Therefore, i70 = i2 = –1
Determine the value of i40
i40 = 1
To find the value of i40, we use the repeating cycle pattern.
We know that the powers of i repeat in a cycle of 4:
i1= i, i2 = –1, i3= – i, i4 = 1
Now, we divide the exponent by 4:
40 ÷ 4 = 10, remainder 0
Here, as the remainder is 0, we can use the pattern:
i40 = (i4)10 = i10 = 1
Check if i20 × i5 = i25
Since i20 × i5 = i25 = i, the given equation is true.
To check if the equation is true, we find i20 and i5:
The powers of i repeat in a cycle of 4, so we write:
i1 = i, i2 = –1, i3 = – i, i4 = 1
We calculate the value of i20:
We divide 20 by 4:
20 ÷ 4 = 5, remainder 0
Here, as the remainder is 0, use the pattern:
I20 = i0 = 1
Now calculate i5:
Dividing 5 by 4:
5 ÷ 4 = 1 with remainder 1
Since the remainder equals 1, we use the pattern:
i5 = i1 = i
Now, multiply i20 and i5
i20 × i5 = 1 × i = i
Similarly, calculate i25
Dividing 25 by 4:
25 ÷ 4 = 6, remainder 1
Since the remainder equals 1, we use the pattern:
i25 = i1 = i
i20 × i5 = i25 = i
Simplify i30 + i12 + i6
i30 + i12 + i6 is simplified to –1.
We find each term in the expression by dividing the exponent by 4:
Finding the value of i30, so dividing 30 by 4
30 ÷ 4 = 7
Since the remainder is 2, we write i30 = i2 = –1
Finding the value of i12, so dividing 12 by 4
12 ÷ 4 = 3
Since the remainder is 0, we write i12 = i0 = 1
Finding the value of i6, so dividing 6 by 4
6 ÷ 4 = 1
Since the remainder is 2, we write i6 = i2 = –1
We substitute these values into the given expression:
(–1) + (1) + (–1) = – 1
Hiralee Lalitkumar Makwana has almost two years of teaching experience. She is a number ninja as she loves numbers. Her interest in numbers can be seen in the way she cracks math puzzles and hidden patterns.
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