A complex number is a number of the form a + bi, where ‘a’ and ‘b’ are real numbers and ‘I’ is an indeterminate satisfying i^{2} = −1. For example, 2 + 3i is a complex number.

The equations x^{2} + 5 = 0, x^{2} + 10 = 0, x^{2} = -1 are not solvable in the real number system i.e, these equations has no real roots. For example, i is the solution of the equation x^{2} = -1 and it has two solutions i.e., x = ± i, where √-1.

The number, ‘I’ is called an imaginary number. Generally, the square root of any negative real number is called imaginary number. The concept of imaginary numbers was first introduced by mathematician “Euler”. He was the one who introduced ‘i’ (read as ‘iota’) to represent √-1. He also defined i2 = -1.

**Multiplication of Two Complex Numbers**

Multiplication of two complex numbers is also a complex number.

In other words, the product of two complex numbers can be expressed in the standard form A + iB where A and B are real.

Let z_{1} = p + iq and z_{2} = r + is be two complex numbers (p, q, r and s are real), then their product z_{1}z_{2 }is defined as

z_{1}z_{2 }= (pr – qs) + i(ps + qr).

**Commutative property of multiplication of two complex numbers:**

For any two complex number z_{1} and z_{2}, we have z_{1}z_{2} = z_{2}z_{1}.

**Proof:**

Let z_{1} = p + iq and z_{2} = r + is, where p, q, r and s are real numbers. Them

z_{1}z_{2} = (p + iq)(r + is) = (pr – qs) + i(ps – rq)

and z_{2}z_{1} = (r + is) (p + iq) = (rp – sq) + i(sp – qr)

= (pr – qs) + i(ps – rq), [Using the commutative of multiplication of real numbers]

Therefore, z_{1}z_{2} = z_{2}z_{1}

Thus, z_{1}z_{2} = z_{2}z_{1 }for all z1, z2 ϵ C.

Hence, the multiplication of complex numbers is commutative on C.

**Examples of commutative property of multiplication of two complex numbers:**

- Show that multiplication of two complex numbers (2 + 3i) and (3 + 4i) is commutative.

Solution:

Let, z_{1} = (2 + 3i) and z_{2} = (3 + 4i)

Now, z_{1}z_{2} = (2 + 3i)(3 + 4i)

= (2 ∙ 3 – 3 ∙ 4) + (2 ∙ 4 + 3 ∙ 3)i

= (6 – 12) + (8 + 9)i

= – 6 + 17i

Again, z_{2}z_{1}= (3 + 4i)(2 + 3i)

= (3 ∙ 2 – 4 ∙ 3) + (3 ∙ 3 + 2 ∙ 4)i

= (6 – 12) + (9 + 8)i

= -6 + 17i

Therefore, z1z2 = z2z1

Thus, z_{1}z_{2} = z_{2}z_{1} for all z_{1}, z_{2} ϵ C.

Hence, the multiplication of two complex numbers (2 + 3i) and (3 + 4i) is commutative.

**Show that multiplication of two complex numbers (3 – 2i) and (-5 + 4i) is commutative.**

Solution:

Let, z_{1} = (3 – 2i) and z2 = (-5 + 4i)

Now, z_{1}z_{2} = (3 – 2i)(-5 + 4i)

= (3 ∙ (-5) – (-2) ∙ 4) + ((-2) ∙ 4 + (-5) ∙ (-2))i

= (-15 – (-8)) + ((-8) + 10)i

= (-15 + 8) + (-8 + 10)i = – 7 + 2i

Again, z_{2}z_{1} = (-5 + 4i)(3 – 2i)

= ((-5) ∙ 3 – 4 ∙ (-2)) + (4 ∙ 3 + (-2) ∙ 4)i

= (-15 + 8) + (12 – 8)i = -7 + 2i

Therefore, z_{1}z_{2} = z_{2}z_{1}

Thus, z_{1}z_{2} = z_{2}z_{1} for all z1, z2 ϵ C.

Hence, the multiplication of two complex numbers (3 – 2i) and (-5 + 4i) is commutative.

Information Source: