Last updated on July 4th, 2025
Imaginary numbers were first introduced by Hero of Alexandria and later developed by Gerolamo Cardano. These numbers are used to solve problems involving the square roots of negative numbers. In this topic, the concept of imaginary numbers and their wider applications will be discussed.
Imaginary numbers are numbers that, when squared, give negative numbers. In other words, they are the square roots of negative numbers. We get an imaginary number by multiplying a non-zero real number by the imaginary unit “i” (iota), where the value of i is √(– 1) or i2 = –1.
For example:
√(–9) = √9 × √(-1) = 3i
√(– 7) = i √7
The square root of a negative number, for any positive real number x:
√(– x) = i√x
For example: √(–36) = i√36 = 6i
A complex number is expressed as the sum of a real number and an imaginary number. It has the form a + ib, where ‘a’ and ‘b’ are real numbers and i is the imaginary number.
A complex number (a + bi) can be represented as a point (a, b) on the Argand plane. For example, we represent the complex number 2 – 4i using the point (2, – 4) on the plane. Since an imaginary number bi can also be expressed as 0 + bi, it is represented as (0, b). For a better understanding, let’s look at an example:
The letter “i” in a complex number (a + ib) is a unique imaginary unit, which is also known as “iota”. Since “i” is not a real number, we represent it on the imaginary axis. Consider ‘i’ as an imaginary unit that moves in a perpendicular direction to the real number line. On the Argand plane, the point (0,1) is used to represent i.
(a + ib) is a complex number where the real part “a” moves along the real axis, and the imaginary part “ib” moves “b” units along the imaginary axis. This implies that a + ib represents the point (a, b) in the plane. Let’s look at the representation of i in math:
To calculate the imaginary numbers, we follow the same methods as real numbers. We will now look at the arithmetic operations performed on imaginary numbers:
Addition and Subtraction of Imaginary Numbers
We can add or subtract imaginary numbers in the same way as combining like terms in algebra. For example:
5i + 4i = 9i
5i – 4i = i
Multiplication of Imaginary Numbers
We multiply imaginary numbers using algebraic multiplication rules and the property i2 = -1.
For example:
5i × 4i = 20i2 = 20(– 1) = – 20
4i2 × (– 3i3) = – 12i5 = – 12(i2)2i = –12 (– 1)2i = –12i
While multiplying complex numbers, apply the following rules to simplify the powers of i:
Here, k is a whole number. This indicates that we can reduce any power of i to one of these four values. Example:
Division of Imaginary Numbers
To divide imaginary numbers, we apply exponent rules:
am/ an = am — n
To rationalize the denominator, we eliminate i by multiplying the numerator and denominator by i; we use the identity 1/i = –i, which can be expanded as:
1/i = 1/i ˟ i/i = i/i2= i /–1 = –i
For example:
6i/3i = 2
Imaginary numbers are abstract concepts, but they are applicable in various real-life situations. Let’s see how they are applied in different fields:
Imaginary numbers are significant units in complex numbers. Understanding them helps students solve problems involving negative square roots. However, students often find them tricky to deal with. Here are a few common mistakes along with tips to avoid them:
Multiply 5i × 6i
–30
We first multiply the coefficients:
5 × 6 = 30
Then, multiply i × i:
Substitute i2 = -1 and combine the terms
30i2 = 30 (–1) = – 30.
Simplify 8/i square
– 8
Substituting i2 = –1 in the denominator:
8/ i2 = 8/ –1 (Since i2 = – 1)
Simplifying the fraction,
8/ –1 = – 8
Determine the value of i to the power 30
i30 = –1
The powers of i repeat in every 4 steps:
i1 = i, i2 = –1, i3 = – i, i4 = 1, and so on.
Here, since the cycle repeats every 4 steps, we divide 30 by 4:
30 ÷ 4 = 7, remainder 2
Now, we use the remainder to determine the value:
We have i2 = –1, so:
i30 = i2 = –1
Simplify √– 64
8i
Since √–1 = i,
We divide √–64 into two parts:
√–64 = √64 × √–1
Since √64 = 8 and √–1 = i,
√–64 = 8i
Calculate (2i)(3 - 4i)
8 + 6i
We apply the distributive property,
2i × 3 = 6i
2i × (-4i) = -8i2
Since i2 = -1, we substitute
-82 = -8(-1) = 8
Now, combining terms, we get,
8 + 6i
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.
: She loves to read number jokes and games.