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 

From Multiplication of Two Complex Numbers to HOME PAGE




Didn't find what you were looking for? Or want to know more information about Math Only Math. Use this Google Search to find what you need.



New! Comments

Have your say about what you just read! Leave me a comment in the box below. Ask a Question or Answer a Question.