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Here we will discuss about the associative property of multiplication of complex numbers.
Commutative property of multiplication complex numbers:
For any three complex numbers z1, z2 and z3, we have (z1z2)z3 = z1(z2z3).
Proof:
Let z1 = a + ib, z2 = c + id and z3 = e + if be any three complex numbers.
Then (z1z2)z3 = {(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)}
= z1(z2z3)
Thus, (z1z2)z3 = z1(z2z3) for all z1, z2, z3 Ο΅ 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 z1 = (2 + 3i), z2 = (4 + 5i) and z3 = (1 + i)
Then (z1z2)z3 = {(2 + 3i)(4 + 5i)}(1 + i)
= (2 β 4 - 3 β 5) + i(2 β 5 + 4 β 3)}(1 + i)
= (8 - 15) + i(10 + 12)}(1 + i)
= (-7 + 22i)(1 + i)
= (-7 β 1 - 22 β 1) + i(-7 β 1 + 1 β 22)
= (-7 β 22) + i(-7 + 22)
= -29 + 15i
Now, z1(z2z3) = (2 + 3i){(4 + 5i)(1 + i)}
= (2 + 3i){(4 β 1 - 5 β 1) + i(4 β 1 + 1 β 5)}
= (2 + 3i){(4 - 5) + i(4 + 5)}
= (2 + 3i)(-1 + 9i)
= {2 β (-1) - 3 β 9} + i{2 β 9 + (-1) β 3}
= (-2 - 27) + i(18 - 3)
= -29 + 15i
Thus, (z1z2)z3 = z1(z2z3) for all z1, z2, z3 Ο΅ C.
Hence, multiplication of complex numbers (2 + 3i), (4 + 5i) and (1 + i) is associative.
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