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
1z
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 |
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 : i
17= i
16 i = i because i
16 is same as i
4
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 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
(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 + i
2(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)
(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
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
From multiplication of complex numbers to Home