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Multiplication of two complex numbers is also a complex number.
In other words, the product of two complex numbers can be expressed in the standard form A + iB where A and B are real.
Let z1 = p + iq and z2 = r + is be two complex numbers (p, q, r and s are real), then their product z1z2 is defined as
z1z2 = (pr - qs) + i(ps + qr).
Proof:
Given z1 = p + iq and z2 = r + is
Now, z1z2 = (p + iq)(r + is) = p(r + is) + iq(r + is) = pr + ips + iqr + i2qs
We know that i2 = -1. Now putting i2 = -1 we get,
= pr + ips + iqr - qs
= pr - qs + ips + iqr
= (pr - qs) + i(ps + qr).
Thus, z1z2 = (pr - qs) + i(ps + qr) = A + iB where A = pr - qs and B = ps + qr are real.
Therefore, product of two complex numbers is a complex number.
Note: Product of more than two complex numbers is also a complex number.
For example:
Let z1 = (4 + 3i) and z2 = (-7 + 6i), then
z1z2 = (4 + 3i)(-7 + 6i)
= 4(-7 + 6i) + 3i(-7 + 6i)
= -28 + 24i - 21i + 18i2
= -28 + 3i - 18
= -28 - 18 + 3i
= -46 + 3i
Properties of multiplication of complex numbers:
If z1, z2 and z3 are any three complex numbers, then
(i) z1z2 = z2z1 (commutative law)
(ii) (z1z2)z3 = z1(z2z3) (associative law)
(iii) z β 1 = z = 1 β z, so 1 acts as the multiplicative identity for the set of complex numbers.
(iv) Existence of multiplicative inverse
For every non-zero complex number z = p + iq, we have the complex number pp2+q2 - iqp2+q2 (denoted by zβ1 or 1z) such that
z β 1z = 1 = 1z β z (check it)
1z is called the multiplicative inverse of z.
Note: If z = p + iq then zβ1 = 1p+iq = 1p+iq β pβiqpβiq = pβiqp2+q2 = pp2+q2 - iqp2+q2.
(v) Multiplication of complex number is distributive over addition of complex numbers.
If z1, z2 and z3 are any three complex numbers, then
z1(z2 + z3) = z1z2 + z1z3
and (z1 + z2)z3 = z1z3 + z2z3
The results are known as distributive laws.
Solved examples on multiplication of two complex numbers:
1. Find the product of two complex numbers (-2 + β3i) and (-3 + 2β3i) and express the result in standard from A + iB.
Solution:
(-2 + β3i)(-3 + 2β3i)
= -2(-3 + 2β3i) + β3i(-3 + 2β3i)
= 6 - 4β3i - 3β3i + 2(β3i)2
= 6 - 7β3i - 6
= 6 - 6 - 7β3i
= 0 - 7β3i, which is the required form A + iB, where A = 0 and B = - 7β3
2. Find the multiplicative inverse of β2 + 7i.
Solution:
Let z = β2 + 7i,
Then Β―z = β2 - 7i and |z|2 = (β2)2 + (7)2 = 2 + 49 = 51.
We know that the multiplicative inverse of z given by
zβ1
= Β―z|z|2
= β2β7i51
= β251 - 751i
Alternatively,
zβ1 = 1z
= 1β2+7i
= 1β2+7i Γ β2β7iβ2β7i
= β2β7i(β2)2β(7i)2
= β2β7i2β49(β1)
= β2β7i2+49
= β2β7i51
= β251 - 751i
11 and 12 Grade Math
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