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

**Commutative property of multiplication complex numbers:**

For any three complex numbers z\(_{1}\), z\(_{2}\) and z\(_{3}\), we have (z\(_{1}\)z\(_{2}\))z\(_{3}\) = z\(_{1}\)(z\(_{2}\)z\(_{3}\)).

**Proof:**

Let z\(_{1}\) = a + ib, z\(_{2}\) = c + id and z\(_{3}\) = e + if be any three complex numbers.

Then (z\(_{1}\)z\(_{2}\))z\(_{3}\) = {(a + ib)(c + id)}(e + if)

= {(ac - bd) +i(ad + cb)}(e + if)

= {(ac - bd)e - (ad + cb)f) + i{(ac - bd)f + (ad + cb)e)

= {a(ce - df) - b(cf + ed)} + i{b(ce - df) + a(ed + cf)

= (a + ib){(cf - df) + i(cf + ed)}

= z\(_{1}\)(z\(_{2}\)z\(_{3}\))

Thus, (z\(_{1}\)z\(_{2}\))z\(_{3}\) = z\(_{1}\)(z\(_{2}\)z\(_{3}\)) for all z\(_{1}\), z\(_{2}\), z\(_{3}\) ϵ C.

Hence, multiplication of complex numbers is associative on C.

Solved example on commutative property of multiplication of
complex numbers:

Show that multiplication of complex numbers (2 + 3i), (4 + 5i) and (1 + i) is associative.

**Solution:**

Let z\(_{1}\) = (2 + 3i), z\(_{2}\) = (4 + 5i) and z\(_{3}\) = (1 + i)

Then (z\(_{1}\)z\(_{2}\))z\(_{3}\) = {(2 + 3i)(4 + 5i)}(1 + i)

= (2 **∙** 4 - 3 **∙** 5) + i(2 **∙** 5 + 4 **∙** 3)}(1 + i)

= (8 - 15) + i(10 + 12)}(1 + i)

= (-7 + 22i)(1 + i)

= (-7 **∙** 1 - 22 **∙** 1) + i(-7 **∙** 1 + 1 **∙** 22)

= (-7 – 22) + i(-7 + 22)

= -29 + 15i

Now, z\(_{1}\)(z\(_{2}\)z\(_{3}\)) = (2 + 3i){(4 + 5i)(1 + i)}

= (2 + 3i){(4 **∙** 1 - 5 **∙** 1) + i(4 **∙** 1 + 1 **∙** 5)}

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

= (2 + 3i)(-1 + 9i)

= {2 **∙** (-1) - 3 **∙** 9} + i{2 **∙** 9 + (-1) **∙** 3}

= (-2 - 27) + i(18 - 3)

= -29 + 15i

Thus, (z\(_{1}\)z\(_{2}\))z\(_{3}\) = z\(_{1}\)(z\(_{2}\)z\(_{3}\)) for all z\(_{1}\), z\(_{2}\), z\(_{3}\) ϵ C.

Hence, multiplication of complex numbers (2 + 3i), (4 + 5i) and (1 + i) is associative.

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