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.
11 and 12 Grade Math
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