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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