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 z1 and z2, we have z1z2 = z2z1.

Proof:

Let z1 = p + iq and z2 = r + is, where p, q, r and s are real numbers. Them

z1z2 = (p + iq)(r + is) = (pr - qs) + i(ps - rq)

and z2z1 = (r + is) (p + iq) = (rp - sq) + i(sp - qr)

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

Therefore, z1z2 = z2z1

Thus, z1z2 = z2z1 for all z1, z2 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, z1 = (2 + 3i) and z2 = (3 + 4i)

Now, z1z2 = (2 + 3i)(3 + 4i)

= (2 3 - 3 4) + (2 4 + 3 3)i

= (6 - 12) + (8 + 9)i

= - 6 + 17i

Again, z2z1 = (3 + 4i)(2 + 3i)

= (3 2 - 4 3) + (3 3 + 2 4)i

= (6 - 12) + (9 + 8)i

= -6 + 17i

Therefore, z1z2 = z2z1

Thus, z1z2 = z2z1 for all z1, 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, z1 = (3 - 2i) and z2 = (-5 + 4i)

Now, z1z2 = (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, z2z1 = (-5 + 4i)(3 - 2i)

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

= (-15 + 8) + (12 - 8)i

= -7 + 2i

Therefore, z1z2 = z2z1

Thus, z1z2 = z2z1 for all z1, z2 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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