Multiplication of Two Complex Numbers

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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