A complex number is a number that includes both a real and an imaginary part. It is written in the form:
z = a + bi
Where:
- a is the real part,
- b is the imaginary part,
- i is the imaginary unit, satisfying i2 = β1.
Algebraic operations on complex numbers follow specific rules based on their real and imaginary parts. The four operations on the complex numbers include:
- Addition
- Subtraction
- Multiplication
- Division
Addition of Complex Numbers
To add two complex numbers, just add the corresponding real and imaginary parts.
(a + bi) + (c + di) = (a + c) + (b + d)iΒ
Examples:
- (7 + 8i) + (6 + 3i) Β = (7 + 6) + (8 + 3)i = 13 + 11i
- (2 + 5i) + (13 + 7i) = (2 + 13) + (7 + 5)i = 15 + 12i
- (-3 - 6i) + (-4 + 14i) = (-3 - 4) + (-6 + 14)i = -7 + 8i
- (4 - 3i ) + ( 6 + 3i) = (4 + 6) + (-3 + 3)i = 10
- (6 + 11i) + (4 + 3i) = (4 + 6) + (11 + 3)i = 10 + 14i
Subtraction of Complex Numbers
To subtract two complex numbers, just subtract the corresponding real and imaginary parts.
(a + bi) β (c + di) = (a β c) + (b β d)iΒ
Examples:
- (6 + 8i) Β - Β (3 + 4i) = (6 - 3) + (8 - 4)i = 3 + 4i
- (7 + 15i) - (2 + 5i) = (7 - 2) + (15 - 5)i = 5 + 10i
- (-3 + 5i) - (6 + 9i) = (-3 - 6) + (5 - 9)i = -9 - 4i
- (14 - 3i) - (-7 + 2i) = (14 - (-7)) + (-3 - 2)i = 21 - 5i
- (-2 + 6i) - (4 + 13i) = (-2 - 4) + (6 - 13)i = -6 - 7i
Multiplication of Two Complex Numbers
Multiplication of two complex numbers is the same as the multiplication of two binomials. Let us suppose that we have to multiply a + bi and c + di. We will multiply them term by term.
(a + bi) β (c + di) = (a + bi) β c + (a + bi) β di
= (a β c + (b β c)i)+((a β d)i + b β d β β1)
= (a β c β b β d + i(b β c + a β d)) Β Β
Example 1: Multiply (1 + 4i) and (3 + 5i).
(1 + 4i) β (3 + 5i) = (3 + 12i) + (5i + 20i2)
= 3 + 17i β 20
= β17 + 17i
Example 2: Multiply 3i and (2 + 6i).
3i β (2 + 6i) can be viewed as (0 + 3i) β (2 + 6i)
= 3i β (2 + 6i)
= 6i + 18i2
= 6i β 18
= β18 + 6i Β Β
Example 3: Multiply (5 + 3i) Β and Β (3 + 4i).
(5 + 3i) β (3 + 4i) = (5 + 3i) β 3 + (5 + 3i) β 4i
= (15 + 9i) + (20i + 12i2)
= (15 β 12) + (20 + 9)i
= 3 + 29i
Note: Multiplication of complex numbers with real numbers or purely imaginary numbers can be done in the same manner.Β
Division of Two Complex NumbersΒ
Division of complex numbers is done by multiplying both the numerator and denominator by the complex conjugate of the denominator.
\frac{a+bi}{c+di} = \frac{(a+bi)(c-di)}{(c+di)(c-di)} \\ \ \\ = \frac{(ac+bd)+ (bc-ad)i}{c^2+d^2} \\ \ \\ = \frac{ac+bd}{c^2+d^2} + \frac{bc-ad}{c^2+d^2}i \\
Example 1:
= \frac{-2}{1+i}*\frac{1-i}{1-i} \\ \ \\ = \frac{-2*(1- i)}{(1+i)*(1 -i)} \\ \ \\ = \frac{-2+2i}{(1+1)+(1- 1)i} \\ \ \\ = \frac{-2+2i}{2} \\ \ \\ = \frac{-2}{2} + \frac{2}{2}i \\ = -1 + iΒ Β Β Β
Example 2:
= \frac{4 + 5i}{2i}*\frac{-2i}{-2i} \\ \ \\ = \frac{(4 + 5i)*-2i}{-4i^2} \\ \ \\ = \frac{10-8i}{4} \\ \ \\ = \frac{10-8i}{4} \\ \ \\ = \frac{10}{4} - \frac{8}{4}i \\ \ \\ = \frac{5}{2} -2iΒ Β Β Β
Example 3:
= \frac{4 + 2i}{-1+i}*\frac{-1-i}{-1-i} \\ \ \\ = \frac{(4 + 2i)*(-1- i)}{(-1+i)*(-1 -i)} \\ \ \\ = \frac{(-4+2)+(-2-4)i}{(1+1)+(1- 1)i} \\ \ \\ = \frac{-2-6i}{2} \\ \ \\ = \frac{-2}{2} - \frac{6}{2}i \\ = -1 -3i
Example 4:
= \frac{2 + 5i}{3+4i}*\frac{3- 4i}{3-4i} \\ \ \\ = \frac{(2 + 5i)*(3- 4i)}{(3+4i)*(3 -4i)} \\ \ \\ = \frac{(6+20)+(15-8)i}{(9+16)+(12- 12)i} \\ \ \\ = \frac{26+7i}{25} \\ \ \\ = \frac{26}{25} + \frac{7}{25}i
Conjugate of a Complex NumberΒ
Two complex numbers, if only the sign of the imaginary part differs then, they are known as a complex conjugate of each other. Thus conjugate of a complex number a + bi would be a - bi.

The complex conjugate of z is denoted by
What's the use of a complex conjugate?
Thus we can observe that multiplying a complex number with its conjugate gives us a real number. Thus the division of complex numbers is possible by multiplying both numerator and denominator with the complex conjugate of the denominator.
Examples of Complex Conjugates
1.
2.
3.
4.
Properties of Complex Conjugates
Property 1:
Proof: Let z = a + ib. Then by definition, (conjugate of z) = a - ib.
Therefore, the conjugate of
Property 2:
Proof: If z1 = a + ib and z2 = c + id then
Now, z1 + z2 = a + ib + c + id = (a + c) + i(b + d)
Therefore,
Property 3:
Proof: If z1 = a + ib and z2 = c + id then
Now, z1 + z2 = a + ib - (c + id) = (a- c) + i(b - d)
Therefore,
= (a - ib) - (c - id)
Property 4:
Proof: If z1 = a + ib and z2 = c + id then
Now, z1 * z2 = a + ib - (c + id) = (ac-bd) + i(bc + ad)
Therefore,
Also,
Property 5:
Proof: According to the problem z2 β 0 β
Let,
z1 = z2*z3