Commutative Property of Multiplication of Complex Numbers

Here we will discuss about the commutative property of multiplication of complex numbers.

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 z\(_{1}\), z\(_{2}\) ϵ C.

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

 

Examples on commutative property of multiplication of two complex numbers:

1. 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, z\(_{1}\)z\(_{2}\) = z\(_{2}\)z\(_{1}\)

Thus, z\(_{1}\)z\(_{2}\) = z\(_{2}\)z\(_{1}\) for all z\(_{1}\), z2 ϵ C.

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

 

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

Solution:

Let, z\(_{1}\) = (3 - 2i) and z\(_{2}\) = (-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 z\(_{1}\), z\(_{2}\) ϵ C.

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

11 and 12 Grade Math 

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