Imaginary Numbers

Last Updated : 01 Dec, 2020

In lower classes, we are taught that the square root of negative numbers can not be taken. However we can take the square root of a negative number, but it involves making use of a new number which is called an imaginary number. So let’s assume some number i exists where: i² = -1. This i is called the imaginary unit. We can observe that we have created a whole new number system (complex numbers), where the square root of i² =-1, and i is called the imaginary unit. Now, complex numbers comprise of real and purely imaginary numbers. We are already familiar with real numbers for eg: 2, 4.03, and π so let’s talk about pure imaginary numbers.

Purely Imaginary Number

A purely imaginary number is a multiple of i. So, -5i+, 27*i are all purely imaginary numbers. They are also called non-real numbers. Thus an imaginary number is a number that can be written as a real number multiplied by the imaginary unit i. Thus complex numbers are of the form a + bi, where a, b are real constants. The complex number a + bi can be broken down into two parts namely

The real part (a)
And the imaginary part (b) [not b*i]

Powers of Imaginary Unit(i)

We know that i²= -1 but what about other powers of i?

$1)\ i = \sqrt{-1} \\ 2)\ i^2 = -1 \\ 3)\ i^3 = i^2*i = -1*i =-i \\ 4)\ i^4 = i^3*i = -i*i = -1*i^2 = 1\\ 5)\ i^5 = i^4*i = 1*i = i \\ 6)\ i^6 = i^5*i = i*i = i^2 = -1 \\ 7)\ i^7 = i^6*i = -1*i = -i \\ 8)\ i^8 = i^7*i = -i*i = -1*i^2 = 1\\$

Do you see the pattern here?

There is a cycle of i, -1, -i , 1 … where every multiple of 4 is 1.
To sum it up , lets say that i is raised to the power of n .
If 1) n mod 4 == 0 [division by 4 leaves 0 as remainder] then the ans is 1 . eg : i4 = i8 = i12 = 1
2) n mod 4 == 1 [division by 4 leaves 1 as remainder] then the ans is i . eg : i = i5 = i9 = i
3) n mod 4 == 2 [division by 4 leaves 2 as remainder] then the ans is -1 . eg : i6 = i10 = -1
4) n mod 4 == 3 [division by 4 leaves 3 as remainder] then the ans is -i . eg : i3 = i7 = -i

Examples:

$1)\ i^{2020} = 1\ (\ as\ 2020\ mod\ 4 == 0\ )\\ Verification :\ i^{2020} = i^{4*505} = {i^4}^{505} = 1^{505} = 1 \\ \ \\ 2)\ i^{105} = i\ (\ as\ 105\ mod\ 4 == 1\ )\\ Verification :\ i^{105} = i^{(4*104)+1} = i*i^{4*104} = i*({i^4}^{104}) = i*(1^{104}) = i \\$

Simplifying Roots of Negative Numbers

Example 1: Suppose you are asked to evaluate the square root of -121.

$\sqrt{-121} \\ = \sqrt{-1*121} \\ = \sqrt{-1} * \sqrt{121} \\ = i*\sqrt{121} \\ =11*i$

Example 2: Suppose you are asked to evaluate the square root of -(1/9).

$\sqrt{-(1/9)} \\ = \sqrt{-1*(1/9)} \\ = \sqrt{-1} * \sqrt{(1/9)} \\ = i*\sqrt{1/9} \\ =1/3*i$

Principal Square Root of Number

The principal Square Root of a non-negative real number is the non-negative square root.

Example:

$\sqrt{9} = +3 ,\ \sqrt{25}= +5 \\ \sqrt{9*25}= \sqrt{9}*\sqrt{25} = 3*5 = 15$

The principal square root of a * b can not be broken down into the principal square root of a * principal square root of b if both a and b are negative.

$\sqrt{121} \\ =\sqrt{-1*-121} \\ \neq \sqrt{-1} * \sqrt{-121}$

Principal Square Root of -1: The imaginary unit is defined as i² = −1. Using this notation, we can think of i as the square root of −1, but we also have (−i)² = i² = −1 and so −i is also a square root of −1. But by convention, the principal square root of −1 is i, or more generally, if x is any non-negative number, then the principal square root of −x is: