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High School Mathematics - 2
2.5 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 binomials governed by the rules of algebra and substituting -1 for i2, we have
z1z2 = (a+ib) (c+id) = ac + iad + ibc + i2bd
= (ac-bd) + i(ad+bc)

Commutative Property: For two complex numbers z1 = a+ib and z2 = c+id, we have
z1z2 = (a+ib) (c+id) = (ac-bd) + i(ad+bc)----1
z2z1 = (c+id) (a+ib) = (ca-db) + i(cb+da)----2
But a, b, c, d are real numbers, therefore
ac-bd = ca-db, ad+bc = cb+da----3
From the above three equations we conclude that z1z2 = z2z1

Associative Property: Consider three complex numbers z1 = a+ib, z2 = c+id and z3 = e+if and if the products (z1z2)z3 and z1(z2z3), we have
(z1z2)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
z1z2(z3) = (a+ib)[(c+id) (e+if)]
(ace-adf-bcf-bde) + i(acf+ade+bce-bdf)----2
From equations 1 and 2 we conclude
z1z2(z3) = (z1z2)z3
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
(ac-bd) + i(ad+bc) = (a+ib)
(ac-bd) = a, ad+bc = b
a(c-1) = bd----1
b(c-1) = -ad----2
Multiplying both sides of equation 1 by a and of 2 by b and adding we get
(a2+b2)(c-1) = 0
Since a2+b2 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
(ac-bd) + i(ad+bc) = 1+i0
ac - bd = 1----1
ad+bc = 0----2
The solution of system of two equations 1 and 2 exist and is given by
c = a/(a2+b2), d = -b/(a2+b2) provided a2+b2 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/(a2+b2)-ib/(a2+b2) = (a-ib) 1/a2+b2) = z/(|z2|

Example: Find the product of -3+-2 and 23-i
Solution: (-3+i 2) (23-i)
Answer: (-6+2)+i(3+26


Directions: Find the product of the following. Also write at least 5 examples of your own.
Q 1: Find the multiplicative inverse of 4-i3
4/25 i
4/25 +3i/25
5i

Q 2: 3i3(15i6)
0+45i
-5i
-45i

Q 3: i-38
1+i
-i
-1+0i

Q 4: i4+i8+i12+i16
4+0i
4-i
3i

Q 5: (-2-i/3)3
-2i/3
-22/3-107/27i
-107/2i

Q 6: (2i)3
8i3
0+8i
0-8i

Q 7: Find the multiplicative inverse of 2+i3
2/7 i
2/7 - i 3/7
7i

Q 8: Find the multiplicative inverse of -i.
-2i
-i
0+i

Question 9: This question is available to subscribers only!

Question 10: This question is available to subscribers only!


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