Multiplication of Complex Numbers
The multiplication of complex numbers: The product of two complex numbers z _{1} = a +ib and z _{2} = c + id is defined as a complex number obtained by the multiplication of these two numbers as binomial governed by the rules of algebra dn substituting -1 for i ^{2} . We have,z _{1} z _{2} = (a + ib)(c + id) = ac + i ad + i bc + i ^{2} bd
= (ac - bd) + i(ad + bc)
For multiplication of complex numbers, the students should know the values of different powers of 'i' . The values of different powers of 'i' are given below.
i | √ -1 |
i^{2} | -1 |
i^{3} | - i |
i^{4} | 1 |
i^{5} | i = √ -1 |
i^{6} | -1 |
i^{7} | -i |
i^{8} | 1 |
For Example : i ^{17} = i ^{16} i = i because i ^{16} is same as i ^{4}
In general, we can say that for any integer ‘k’
i^{4k} = 1 , i i^{4k +1 } = i i^{4k+ 2} = -1 , i^{4k+3} = -i |
Properties of multiplication of complex numbers
Closure : The product of two complex numbers is , by definition , a complex number. Hence, the set of complex numbers is closed under multiplication.Commutative property : For two complex numbers z _{1} = a + ib and
z _{2} = c + id , we have
z _{1} . z _{2} = (a + ib)(c + id) = (ac -bd) + i(ad + bc) (since i ^{1} = -1
z _{2} . z _{1} = (c + id)(a + ib) = (ca-bd) + i(cb + da)
But a, b, c , d are real numbers, so, ac - bd = ca - db and ad + bc = cb + da
Hence, multiplication of complex-numbers is commutative.
Associative Property : Consider the three complex numbers,
z _{1} = a + ib , z _{2} = c + id and z _{3} = e + if
(z_{1} . z_{2} ). z_{3} = [(a + ib).( c + id )] .(e + if)
=[(ac - bd)+ i(ad +bc)] . (e + if)
=(ac-bd). e + i(ad +bc)e + i(ac -bd)f + i ^{2} (ad +bc).f
=(ace -bde - adf -bcf) + i(ade + bce + acf -bdf) -------------(1)
z_{1}.(z_{2} . z_{3}) = (a+ib).[(c +id).(e +if)]
= (ace - adf - bcf -bde) + i(acf + ade + bce -bdf) -----(2)
Thus, from (1) and (2)
(z _{1} . z _{2} ). z _{3} = z _{1} .(z _{2} . z _{3} )
Multiplication Identity: Let c + id be the multiplicative identity of a + ib. Then
(a + ib)(c + id) = a + ib
⇒ (ac - bd) + i(ad + bc) = a + ib
⇒ ac - bd = a and ad + bc = b
ac - a = bd and bc - b = -ad
a(c - 1) = bd ----(1)
b(c - 1) = -ad ----(2)
Multiply equation (1) by a and equation(2) by b and then add
a ^{2} (c - 1) = abd
b ^{2} (c - 1) = -abd
------------------------------
(a ^{2} + b ^{2} )(c - 1) = 0
so either a ^{2} + b ^{2} = 0 or c - 1 = 0
but a ^{2} + b ^{2} ≠ 0
so c - 1 = 0 ⇒ c = 1
∴ d = 0
c + id = 1 + i0 = 1
Hence the multiplicative identity of the complex number is 1.
Multiplicative inverse: A complex number 'w' is called the multiplicative inverse of complex number z, if z . w = 1. The multiplicative inverse is denoted by z
FOIL
( 4 x 2)(4 x 12i)(2i x 2)(2i x 12i) = + + +
8(-24) = - 16 + 52 i
^{2}^{2} = 9 - 4i
2) Find the multiplicative inverse of - 3 + 4i Let z = -3 + 4i , then z = z̄ /|z|