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 z1 = p + iq and z2 = r + is be two complex numbers (p, q, r and s are real), then their product z1z2 is defined as

z1z2 = (pr - qs) + i(ps + qr).

Proof:

Given z1 = p + iq and z2 = r + is

Now, z1z2 = (p + iq)(r + is) = p(r + is) + iq(r + is) = pr + ips + iqr + i2qs

We know that i2 = -1. Now putting i2 = -1 we get,

= pr + ips + iqr - qs

= pr - qs + ips + iqr

= (pr - qs) + i(ps + qr).

Thus, z1z2 = (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 z1 = (4 + 3i) and z2 = (-7 + 6i), then

z1z2 = (4 + 3i)(-7 + 6i)

= 4(-7 + 6i) + 3i(-7 + 6i)

= -28 + 24i - 21i + 18i2

= -28 + 3i - 18

= -28 - 18 + 3i

= -46 + 3i

 

Properties of multiplication of complex numbers:

If z1, z2 and z3 are any three complex numbers, then

(i) z1z2 = z2z1 (commutative law)

(ii) (z1z2)z3 = z1(z2z3) (associative law)

(iii) z βˆ™ 1 = z = 1 βˆ™ 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 pp2+q2 - iqp2+q2 (denoted by zβˆ’1 or 1z) such that

z βˆ™ 1z = 1 = 1z βˆ™ z (check it)

1z is called the multiplicative inverse of z.

Note: If z = p + iq then zβˆ’1 = 1p+iq = 1p+iq βˆ™ pβˆ’iqpβˆ’iq = pβˆ’iqp2+q2 = pp2+q2 - iqp2+q2.

(v) Multiplication of complex number is distributive over addition of complex numbers.

If z1, z2 and z3 are any three complex numbers, then

z1(z2 + z3) = z1z2 + z1z3

and (z1 + z2)z3 = z1z3 + z2z3

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 ¯z = √2 - 7i and |z|2 = (√2)2 + (7)2 = 2 + 49 = 51.

We know that the multiplicative inverse of z given by

zβˆ’1

= Β―z|z|2

= √2βˆ’7i51

= √251 - 751i

Alternatively,

zβˆ’1 = 1z

= 1√2+7i

= 1√2+7i Γ— √2βˆ’7i√2βˆ’7i

= √2βˆ’7i(√2)2βˆ’(7i)2

= √2βˆ’7i2βˆ’49(βˆ’1)

= √2βˆ’7i2+49

= √2βˆ’7i51

= √251 - 751i





11 and 12 Grade Math 

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