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 z\(_{1}\) = p + iq and z\(_{2}\) = r + is be two complex numbers (p, q, r and s are real), then their product z\(_{1}\)z\(_{2}\) is defined as
z\(_{1}\)z\(_{2}\) = (pr - qs) + i(ps + qr).
Proof:
Given z\(_{1}\) = p + iq and z\(_{2}\) = r + is
Now, z\(_{1}\)z\(_{2}\) = (p + iq)(r + is) = p(r + is) + iq(r + is) = pr + ips + iqr + i\(^{2}\)qs
We know that i\(^{2}\) = -1. Now putting i\(^{2}\) = -1 we get,
= pr + ips + iqr - qs
= pr - qs + ips + iqr
= (pr - qs) + i(ps + qr).
Thus, z\(_{1}\)z\(_{2}\) = (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 z\(_{1}\) = (4 + 3i) and z\(_{2}\) = (-7 + 6i), then
z\(_{1}\)z\(_{2}\) = (4 + 3i)(-7 + 6i)
= 4(-7 + 6i) + 3i(-7 + 6i)
= -28 + 24i - 21i + 18i\(^{2}\)
= -28 + 3i - 18
= -28 - 18 + 3i
= -46 + 3i
Properties of multiplication of complex numbers:
If z\(_{1}\), z\(_{2}\) and z\(_{3}\) are any three complex numbers, then
(i) z\(_{1}\)z\(_{2}\) = z\(_{2}\)z\(_{1}\) (commutative law)
(ii) (z\(_{1}\)z\(_{2}\))z\(_{3}\) = z\(_{1}\)(z\(_{2}\)z\(_{3}\)) (associative law)
(iii) z ∙ 1 = z = 1 ∙ 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 \(\frac{p}{p^{2} + q^{2}}\) - i\(\frac{q}{p^{2} + q^{2}}\) (denoted by z\(^{-1}\) or \(\frac{1}{z}\)) such that
z ∙ \(\frac{1}{z}\) = 1 = \(\frac{1}{z}\) ∙ z (check it)
\(\frac{1}{z}\) is called the multiplicative inverse of z.
Note: If z = p + iq then z\(^{-1}\) = \(\frac{1}{p + iq}\) = \(\frac{1}{p + iq}\) ∙ \(\frac{p - iq}{p - iq}\) = \(\frac{p - iq}{p^{2} + q^{2}}\) = \(\frac{p}{p^{2} + q^{2}}\) - i\(\frac{q}{p^{2} + q^{2}}\).
(v) Multiplication of complex number is distributive over addition of complex numbers.
If z\(_{1}\), z\(_{2}\) and z\(_{3}\) are any three complex numbers, then
z\(_{1}\)(z\(_{2}\) + z3) = z\(_{1}\)z\(_{2}\) + z\(_{1}\)z\(_{3}\)
and (z\(_{1}\) + z\(_{2}\))z\(_{3}\) = z\(_{1}\)z\(_{3}\) + z\(_{2}\)z\(_{3}\)
The results are known as distributive laws.
Solved examples on multiplication of two complex numbers:
1. Find the product of two complex numbers (-2 + √3i) and (-3 + 2√3i) and express the result in standard from A + iB.
Solution:
(-2 + √3i)(-3 + 2√3i)
= -2(-3 + 2√3i) + √3i(-3 + 2√3i)
= 6 - 4√3i - 3√3i + 2(√3i)\(^{2}\)
= 6 - 7√3i - 6
= 6 - 6 - 7√3i
= 0 - 7√3i, which is the required form A + iB, where A = 0 and B = - 7√3
2. Find the multiplicative inverse of √2 + 7i.
Solution:
Let z = √2 + 7i,
Then \(\overline{z}\) = √2 - 7i and |z|\(^{2}\) = (√2)\(^{2}\) + (7)\(^{2}\) = 2 + 49 = 51.
We know that the multiplicative inverse of z given by
z\(^{-1}\)
= \(\frac{\overline{z}}{|z|^{2}}\)
= \(\frac{√2 - 7i}{51}\)
= \(\frac{√2}{51}\) - \(\frac{7}{51}\)i
Alternatively,
z\(^{-1}\) = \(\frac{1}{z}\)
= \(\frac{1}{√2 + 7i }\)
= \(\frac{1}{√2 + 7i }\) × \(\frac{√2 - 7i}{√2 - 7i }\)
= \(\frac{√2 - 7i}{(√2)^{2} - (7i)^{2}}\)
= \(\frac{√2 - 7i}{2 - 49(-1)}\)
= \(\frac{√2 - 7i}{2 + 49}\)
= \(\frac{√2 - 7i}{51}\)
= \(\frac{√2}{51}\) - \(\frac{7}{51}\)i
11 and 12 Grade Math
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