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 binomials governed by the rules of algebra and substituting 1 for i^{2}, we have
z_{1}z_{2} = (a+ib) (c+id) = ac + iad + ibc + i_{2}bd
= (acbd) + i(ad+bc)
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) = (acbd) + i(ad+bc)1
z_{2}z_{1} = (c+id) (a+ib) = (cadb) + i(cb+da)2
But a, b, c, d are real numbers, therefore
acbd = cadb, ad+bc = cb+da3
From the above three equations we conclude that z_{1}z_{2} = z_{2}z_{1}
Associative Property: Consider three complex numbers z_{1} = a+ib, z_{2} = c+id and z_{3} = e+if and if the products (z_{1}z_{2})z_{3} and z_{1}(z_{2}z_{3}), we have
(z_{1}z_{2})z_{3} = [(a+ib) (c+id)](e+if)
= [(acbd)+i(ad+bc)](e+if)
= (acbd)e + i(ad+bc)e + i(acbd)f + i^{2}(ad+bc)f
(acebdeadfbcf) + i(ade+bce+acfbdf)1
z_{1}z_{2}(z_{3}) = (a+ib)[(c+id) (e+if)]
(aceadfbcfbde) + i(acf+ade+bcebdf)2
From equations 1 and 2 we conclude
z_{1}z_{2}(z_{3}) = (z_{1}z_{2})z_{3}
Hence associative law of multiplication holds for complex numbers.
Multiplicative Identity: Let c+id be the multiplicative identity of a+ib, then
(a+ib)(c+id) = a+ib
(acbd) + i(ad+bc) = (a+ib)
(acbd) = a, ad+bc = b
a(c1) = bd1
b(c1) = ad2
Multiplying both sides of equation 1 by a and of 2 by b and adding we get
(a^{2}+b^{2})(c1) = 0
Since a^{2}+b^{2} not equal to 0, so c = 1
Therefore d = 0
Thus c+id = 1+i0 = 1
Hence the multiplicative identity is the complex number 1+i0, written simply as 1, i.e 1.(a+ib) = (a+ib).1 = a+ib
Multiplicative Inverse: A complex number w is called the multiplicative inverse of z if zw = 1. The multiplicative inverse, w, of z is denoted by z^{1}.
Let z = a+ib be a complex number and w = c+id be its multiplicative inverse.
Then
zw = 1
(a+ib) (c+id) = 1+i0
(acbd) + i(ad+bc) = 1+i0
ac  bd = 11
ad+bc = 02
The solution of system of two equations 1 and 2 exist and is given by
c = a/(a^{2}+b^{2}), d = b/(a^{2}+b^{2}) provided a^{2}+b^{2} not equal to 0.
We, therefore , observe that the multiplicative inverse w(=z^{1}) of any complex number z = a+ib not equal to zero exist and is given by
z^{1} = a/(a^{2}+b^{2})ib/(a^{2}+b^{2}) = (aib) 1/a^{2}+b^{2}) = z/(z^{2}
Example: Find the product of Ö3+Ö2 and 2Ö3i
Solution: (Ö3+i Ö2) (2Ö3i)
Answer: (6+Ö2)+i(Ö3+2Ö6
Directions: Find the product of the following. Also write at least 5 examples of your own.
