Multiplication of Complex Numbers

The multiplication of complex numbers: The product of two complex numbers z₁= a +ib and z₂ = 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 i². We have,
z₁z₂ = (a + ib)(c + id) = ac + i ad + i bc + i² 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
i2	-1
i3	- i
i4	1
i5	i = √ -1
i6	-1
i7	-i
i8	1

From the above table you can see that 1st four powers of ‘i’ are all different but then there is a repetition thereafter, in cycles of four.
For Example : i¹⁷= i¹⁶ i = i because i¹⁶ is same as i⁴
In general, we can say that for any integer ‘k’

i4k = 1 ,    i     i4k +1 = i  i4k+ 2 = -1 ,     i4k+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 z₁= a + ib and
z₂= c + id , we have
z₁. z₂ = (a + ib)(c + id) = (ac -bd) + i(ad + bc) (since i¹= -1
z₂. z₁ = (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,
z₁ = a + ib , z₂ = c + id and z₃ = e + if
(z₁ . z₂ ). z₃ = [(a + ib).( c + id )] .(e + if)
=[(ac - bd)+ i(ad +bc)] . (e + if)
=(ac-bd). e + i(ad +bc)e + i(ac -bd)f + i²(ad +bc).f
=(ace -bde - adf -bcf) + i(ade + bce + acf -bdf) -------------(1)
z₁.(z₂ . z₃) = (a+ib).[(c +id).(e +if)]
= (ace - adf - bcf -bde) + i(acf + ade + bce -bdf) -----(2)
Thus, from (1) and (2)
(z₁ . z₂ ). z₃ = z₁.(z₂ . z₃)

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
a²(c - 1) = abd
b² (c - 1) = -abd
------------------------------
(a² + b² )(c - 1) = 0
so either a² + b² = 0 or c - 1 = 0
but a² + b² ≠ 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

Examples on multiplication of complex numbers
1) Multiply the following:
a) (4 + 2i) (2 + 12i)
Here we will use a FOILmethod
= ( 4 x 2) + (4 x 12i) + (2i x 2) + (2i x 12i)
= 8 + 48i + 4i + (24i ²)
= 8 + 52 i + (-24)
= - 16 + 52 i

b) (3 + 2i)(3 - 2i)
= (3)² - (2i)²
= 9 - 4i²
= 9 + 4
= 13

2) Find the multiplicative inverse of - 3 + 4i
Solution: Let z = -3 + 4i , then
z ^-1 = z̄ /|z|²
= (-3 - 4i)/(9 + 16)
= -3/25 - 4i/25

11th grade math

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