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 
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’
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) = (cabd) + i(cb + da)
But a, b, c , d are real numbers, so,
ac  bd = ca  db and ad + bc = cb + da
Hence, multiplication of complexnumbers 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)
=(acbd). 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
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  4i^{2}
= 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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