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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