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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