How to Find the Modulus of a Complex Number

Here you will learn how to find the modulus of a complex number and properties of modulus with examples.

Let’s begin – 

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 – 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 – 4i, \(z_2\) = -5 + 2i and \(z_3\) = 1 + \(\sqrt{-3}\)

| \(z_1\) | = | 3 – 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, then 

(i) | z | = 0 \(\iff\) z = 0  i.e.  Re (z) = Im (z) = 0

(ii) | z | = | \(\bar{z}\) | = | -z |

(iii) – | z | \(\le\) Re (z) \(\le\) | z |  ;  – | z | \(\le\) Im (z) \(\le\) | z | 

(iv) \(z\bar{z}\) = \(| z |^2\)

(v) | \(z_1 z_2\) | = | \(z_1\) | | \(z_2\) |

(vi) | \(z_1\over z_2\) | = \( | z_1 | \over | z_2 |\) ,  \(z_2\) \(\ne\) 0

(vii)  \( | z_1 + z_2 |^2\) = \( | z_1 |^2\) + \( | z_2 |^2\) + \( 2 Re (z_1\bar{z_2})\)

(viii)  \( | z_1 – z_2 |^2\) = \( | z_1 |^2\) + \( | z_2 |^2\) – \( 2 Re (z_1\bar{z_2})\)

(ix)  \( | z_1 + z_2 |^2\) + \( | z_1 – z_2 |^2\) = 2(\( | z_1 |^2\) + \( | z_2 |^2\))

(x) \( | az_1 – bz_2 |^2\) + \( | bz_1 + az_2 |^2\) = \(a^2 + b^2\) (\( | z_1 |^2\) + \( | z_2 |^2\)), where a. b \(\in\) R.

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