Here you will learn how to find the modulus of a complex number and properties of modulus with examples.
Letโs begin โย
How to Find the Modulus of a Complex Number
The modulus of a complex number z = a + ib is denoted by | z | and is defined as
| z | = \(\sqrt{a^2 + b^2}\) = \(\sqrt{[Re (z)]^2 + [Im (z)]^2}\)
Clearly, | z | \(\ge\) 0 for all z \(\in\) C.
Example : If \(z_1\) = 3 โ 4i, \(z_2\) = -5 + 2i and \(z_3\) = 1 + \(\sqrt{-3}\), then find modulus of \(z_1\), \(z_2\) and \(z_3\).
Solution : We have, \(z_1\) = 3 โ 4i, \(z_2\) = -5 + 2i and \(z_3\) = 1 + \(\sqrt{-3}\)
| \(z_1\) | = | 3 โ 4i | = \(\sqrt{3^2 + (-4)^2}\) = 5,
| \(z_2\) | = | 5 + 2i | = \(\sqrt{(-5)^2 + 2^2}\) = \(\sqrt{29}\)
and, | \(z_3\) | = | 1 + \(\sqrt{-3}\) | = \(\sqrt{1^2 + (\sqrt{3})^2}\) = 2
Remark : In the set C of all complex numbers, the order relation is not defined. As such \(z_1\) > \(z_2\) or \(z_1\) < \(z_2\) has no meaning but | \(z_1\) | > | \(z_2\) | or | \(z_1\) | < | \(z_2\) | has got its meaning since | \(z_1\) | and | \(z_2\) | are real numbers.
Properties of Modulus
If z, \(z_1\), \(z_2\) \(\in\) C, thenย
(i) | z | = 0 \(\iff\) z = 0ย i.e.ย Re (z) = Im (z) = 0
(ii) | z | = | \(\bar{z}\) | = | -z |
(iii) โ | z | \(\le\) Re (z) \(\le\) | z |ย ;ย โ | z | \(\le\) Im (z) \(\le\) | z |ย
(iv) \(z\bar{z}\) = \(| z |^2\)
(v) | \(z_1 z_2\) | = | \(z_1\) | | \(z_2\) |
(vi) | \(z_1\over z_2\) | = \( | z_1 | \over | z_2 |\) ,ย \(z_2\) \(\ne\) 0
(vii)ย \( | z_1 + z_2 |^2\) = \( | z_1 |^2\) + \( | z_2 |^2\) + \( 2 Re (z_1\bar{z_2})\)
(viii)ย \( | z_1 โ z_2 |^2\) = \( | z_1 |^2\) + \( | z_2 |^2\) โ \( 2 Re (z_1\bar{z_2})\)
(ix)ย \( | z_1 + z_2 |^2\) + \( | z_1 โ z_2 |^2\) = 2(\( | z_1 |^2\) + \( | z_2 |^2\))
(x) \( | az_1 โ bz_2 |^2\) + \( | bz_1 + az_2 |^2\) = \(a^2 + b^2\) (\( | z_1 |^2\) + \( | z_2 |^2\)), where a. b \(\in\) R.
