Imaginary Numbers

Imaginary numbers are numbers that are not real. We know that the quadratic equation is of the form ax2 + bx + c = 0, where the discriminant is b2 – 4ac. Whenever the discriminant is less than 0, finding square root becomes necessary for us. For this, we assume the root of -1 = i. The notation “i” is the foundation for all imaginary numbers. The solution written by using this imaginary number in the form a+bi is known as a complex number. In other words, a complex number is one which includes both real and imaginary numbers.

Complex Number

Complex numbers are the combination of both real numbers and imaginary numbers. The complex number is of the standard form: a + bi


a and b are real numbers

i is an imaginary unit.

Real Numbers Examples : 3, 8, -2, 0, 10

Imaginary Number Examples: 3i, 7i, -2i, √A

Complex Numbers Examples: 3 + 4 i , 7 – 13.6 i , 0 + 25 i = 25 i , 2 + i.

Imaginary Number Rules

Consider an example, a+bi is a complex number. For a +bi, the conjugate pair is a-bi. The complex roots exist in pairs so that when multiplied, it becomes equations with real coefficients.

Consider the pure quadratic equation: x 2 = a, where a is a known value. Its solution may be presented as x = √a. Therefore, the rules for some imaginary numbers are:

  • i = √1
  • i2 = -1
  • i3 = -i
  • i4 = +1
  • i4n = 1
  • i4n-1= -i

Operations on Imaginary Numbers

The basic arithmetic operations in Mathematics are addition, subtraction, multiplication, and division. Let us discuss these operations on imaginary numbers.

Let us assume the two complex numbers: a + bi and c + di.

Addition of Numbers Having Imaginary Numbers

When two numbers, a+bi, and c+di are added, then the real parts are separately added and simplified, and then imaginary parts separately added and simplified. Here, the answer is (a+c) + i(b+d).

Subtraction of Numbers Having Imaginary Numbers

When c+di is subtracted from a+bi, the answer is done like in addition. It means, grouping all the real terms separately and imaginary terms separately and doing simplification. Here, (a+bi)-(c+di) = (a-c) +i(b-d).

Multiplication of Numbers Having Imaginary Numbers

Consider (a+bi)(c+di)

It becomes

(a+bi)(c+di) = (a+bi)c + (a+bi)di

= ac+bci+adi+bdi2

= (ac-bd)+i(bc+ad)

Division of Numbers Having Imaginary Numbers

Consider the division of one imaginary number by another.

(a+bi) / ( c+di)

Multiply both the numerator and denominator by its conjugate pair, and make it real. So, it becomes

(a+bi) / ( c+di) = (a+bi) (c-di) / ( c+di) (c-di) = [(ac+bd)+ i(bc-ad)] / c2 +d2.

Imaginary Numbers Example


Solve the imaginary number i7


The given imaginary number is i7

Now, split the imaginary number into terms, and it becomes

i7 = i2 × i2 × i2 × i

i7 = -1 × -1 × -1 × i

i7 = -1 × i

i7 = – i

Therefore, i7 is – i.

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