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****From Commutative Property of Multiplication of Complex Numbers** **to HOME PAGE**

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