# 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